m 


UNIVERSITY  OF  CALIFORNIA 
AT  LOS  ANGELES 


.- 


• 


JV 

LI 


BOLECTIC    EDUCATIONAL    COURHS. 


RAY'S    HIGHER    ARITHMETIC. 


THE    PRINCIPLES 

or 

ARITHMETIC, 


ANALYZED  AND  PRACTICALLY 


FOR    ADVANCED    STUDENTS. 


BY    JOSEPH    RAY,    M.:f>;    < 

LATE    PROFESSOR    OK    MATHEMATICS    IN    WOODWARD    COLLKOA. ' 


EDITED  BY  CHAS.  E.  MATTHEWS,  M.' 


REVISED    STEREOTYPE    EDITION.    ., 


CINCINNATI: 
WILSON,    HINKLE    &    GO 

PHIL'A :  CLAXTON,  REMSEN  &  HAFFELFINGER. 
NEW  YORK:  CLARK  &  MAYNARD. 


146022 


THE    BEST    AND    CHEAPEST 

MATHEMATICAL  WORKS. 

BY  PROFESSOR  JOSEPH  RAY. 


Each  BOOK  of  the  Arithmetical  Course,  as  well  as  the  Algebraic,  is 
Complete  in  itself,  and  is  sold  separately. 

PRIMARY  ARITHMETIC.— RAY'S   ARITHMETIC,  FIRST   BOOK: 

simple  mental  Lessons  and  tables  for  little  learners. 
NTELLECTTIAL    ARITHMETIC.—  RAY'S   ARITHMETIC,    SECOND 

BOOK  :  the  most  interesting  Intellectual  Arithmetic  extant. 
PRACTICAL  ARITHMETIC.— RAY'S  ARITHMETIC,  THIRD  BOOK: 

for  schools  and  academies  ;  a  full  and  complete  treatise,  on  the  in- 
ductive and  analytical  methods  of  instruction. 
KEY  TO  By&."#'S  ARITHMETIC,  containing  solutions  to  the  questions ; 

also,of;te,si  Examples  for  the  Slate  and  Blackboard. 
RAY'S 'HIGHER  ARITHMETIC.— The  Principles  of  Arithmetic,  ana- 

lysad  aoel  p'ractically  applied.     For  advanced  classes. 
ELEMENTARY  ALGEBRA.— RAY'S  ALGEBRA,  FIRST  BOOK ;  for 

cbtemo^«6cliools  and  academies;  a  simple,  progressive,  and  thorough 

elementary  .treatise. 
HIQHER  ifcLXJEBRA.— RAY'S  ALGEBRA,  SECOND  BOOK  ;  for  ad- 

v:-:n<!edv'8ttidents   in  academies,  and  for  colleges;    a  progressive, 

IftWd  anU* Comprehensive  work. 
KEY  TO  IIAY'S  ALGEBRA,  FIRST  AND  SECOND  BOOKS.  In  one  vol. 


ring  for  |ubluati0n, 


L-TTHE.  ELEMENTS  OF  GEOMETRY,  embracing  plain  and  solid 
geometry^  with  numerous  practical  exercises,  n.  —  TRIGONOMETRY 
ANi}  iMEJnURATION,  containing  logarithmic  computations,  plane 
and  'spherical  trigonometry,  with  their  applications,  mensuration  of 
plaues>  and  solids,  with  logarithmic  and  other  tables.  III.  —  SUR- 
VEYING ,'AJSTi)  NAVIGATION  ;  surveying  and  leveling,  navigation, 
barftm'ttrio  hights,  &c. 

Tei.Ve  ffefluwed  by  others,  forming  a  complete  Mathematical  Course 
for'Sahools.nei'd  colleges. 

CHAS.  E.'  MATTHEWS,  M.  A.  for  many  years  a  pupil  of,  and  subse- 
quently an  'associate  Instructor  with  the  late  DR.  RAY  in  the  Mathe- 
matical department  of  Woodward  College,  will  edit  and  superintend 
the  publication  of  the  unfinished  parts  of  the  series. 

ENTERED  according  to  Act  of  Congress,  in  the  year  Eighteen  Hundred 
»nd  Fifty  Six,  by  WINTHROP  B.  SMITH,  in  t'^e  Clerk's  Office  of  the  District 
Court  of  the  United  States  for  the  Southern  District  of  Ohio. 


fltereotyped  by  C.  F.  O'Driscoll  4  Co. 


Q.A 
\  03 


PREFACE. 


As  my  name  appears  in  connection  with  that  of  the  late  I'K. 
JOSEPH  RAY,  on  the  title-page  of  this  work,  it  is  proper  to  account 
for  the  circumstance  by  referring  briefly  to  matters  personal 
in  their  nature.  As  an  author,  Dr.  Ray  is  known  to  the  public 
at  large,  through  the  medium  of  his  excellent  mathematical 
publications.  As  an  able  and  faithful  teacher,  his  merits  aro 
deeply  impressed  on  those  who  have  been  so  fortunate  as  to  come 
within  reach  of  his  immediate  instructions  and  example.  In 
every  line  of  duty  he  was  conspicuous  for  unremitting  industry, 
and  in  all  the  relations  of  life,  his  first  desire  was  to  be  of  service 
to  others. 

To  me,  his  memory  is  endeared  by  many  acts  of  friendship  and 
confidence,  extending  over  a  period  of  twenty  years,  sixteen  of 
which  were  spent  with  him  as  pupil  and  colleague,  in  Woodward 
College  and  High  School. 

In  his  last  illness,  Dr.  Ray  expressed  a  wish  that  I  should  pre- 
pare for  publication,  his  Higher  Arithmetic,  then  unfinished,  and 
directed  hia  manuscripts  and  materials  to  be  handed  to  me  for 
chat  purpose. 

In  performing  the  part  thus  assigned  to  me,  I  have  endeavored 
to  preserve  the  spirit  and  pursue  the  course  indicated  in  the  other 
Arithmetics  of  this  series.  With  this  view,  the  language  of  the 
author  has  been  retained  without  material  change  as  far  as  the 
manuscript  was  complete  and  ready  for  publication.  Whenever 
a  change  has  been  made,  it  has  been  done  with  great  caution 
and  with  an  anxious  desire  to  carry  out  the  general  plan  of 
the  work.  New  matter  has  been  occasionally  introduced  when  it 
deemed  necessary  to  render  a  demonstration  more  clear  and  con- 
clusive, or  the  treatment  of  a  subject  more  full  and  satisfactory. 

This  has  been  especially  the  case  in  the  Contractions  in  Multi- 
plication, Contractions  in  Division,  Greatest  Common  Divisor  and 
Least  Common  Multiple  both  of  whole  numbers  and  fractions 

fiii) 


IT  PREFACE. 

Pure  Arithmetic,  comprising  Simple  Numbers,  Common  and 
Decimal  Fractions  has  been  discussed  before  any  of  ita  applica- 
tions. This  arrangement  is  philosophical,  and  not  open  to  objec- 
tion in  <i  \vork  of  this  character. 

In  questions  of  proportion,  and  generally  throughout  the  book, 
the  analytic  method  of  solution  has  been  preferred  to  mere  for- 
mal and  irrational  directions ;  for  no  true  development  of  the 
intellectual  powers  or  satisfactory  knowledge  of  any  science  can 
be  attained,  unless  the  spirit  of  every  operation  is  clearly  seen 
through  its  form. 

"Where  the  importance  of  the  subject  seemed  to  demand  it,  as 
in  Insurance,  Simple  Interest,  Compound  Interest  and  their  appli- 
cations, brief  methods  of  operation  have  been  given,  and  prao 
tical  rules  deduced  for  the  use  of  accountants  and  men  of  business. 

The  difficulties  and  obscurities  attending  some  of  the  applica- 
tions of  Simple  and  Compound  Interest  are  inseparable  from  the 
subjects  themselves,  and  it  is  better  to  meet  and  overcome  these 
obstacles  in  the  school-room  than  in  the  counting-room.  No 
apology  is  therefore  offered  for  introducing  these  subjects  and 
discussing  them  at  large. 

The  exercises  are  numerous,  most  of  them  new  and  interesting, 
and  have  been  prepared  with  a  view  to  practical  utility.  AVhile 
they  afford  a  full  and  thorough  exercise  in  the  principles  of  Arith- 
metic, at  the  same  time,  they  enable  the  pupil  to  make  use  of  his 
own  knowledge  to  the  best  advantage. 

Circulating  Decimals  and  other  matters  rather  curious  than  ex- 
tensively useful,  may  bo  omitted  until  a  review,  if  the  want  of  time 
or  the  character  of  the  pupils  make  it  necessary ;  and  any  exam- 
ples considered  too  difficult  at  first  may  also  be  postponed  in  the 
farm)  way 

THE  EDITOR 

STEREOTYPE   EDITION. 

The  marked  approbation  extended  to  this  volume,  and  its  wide  in- 
troduction into  the  best  Schools  of  the  country,  having  led  to  the  sala 
of  several  editions  in  a  few  months,  the  work  has  been  thoroughly 
revised,  and  is  now  presented  in  a  permanent  stereotype  form.  In  th« 
revision,  by  a  careful  abridgment  of  language,  the  ideas  are  pre- 
eented  with  greater  precision  and  perspicuity,  the  book  condensed, 
wrid  a  large  number  of  new  and  interesting  examples  added. 


CONTENTS. 


•Mi 

INTRODUCTION 9 

SIMPLE  NUMBERS. 

Numeration  and  Notation 11 

Addition 19 

Subtraction 22 

Multiplication 26 

Division 35 

General  Principles  of  Multiplication  and  Division 47 

Summary  of  Principles 51 

PROPERTIES  OF  NUMBERS 55 

Factoring 66 

Greatest  Common  Divisor 60 

Least  Common  Multiple 64 

Proofs  of  the  Rules  by  casting  out  the  9's  and  ll's 67 

Cancellation 71 

COMMON  TRACTIONS 72 

Numeration  and  Notation  of  Fractions 74 

Reduction  of  Fractions 77 

Addition  of  Fractions 84 

Subtraction  of  Fractions 85 

Multiplication  of  Fractions 87 

Division  of  Fractions 89 

To  Reduce  Complex  Fractions  to  Simple  ones 92 

To  find  the  Greatest  Common  Divisor  of  Fractions 93 

To  find  the  Least  Common  Multiple  of  Fractions U4 

DECIMAL  FRACTIONS 97 

Numeration  and  Notation  of  Decimals 99 

Reduction  of  Decimals 103 

Addition  of  Decimals 107 

Subtraction  of  Decimals 108 

Multiplication  of  Decimals 110 

Division  of  Decimals .' 114 

CIRCULATING  DECIMALS 120 

Reduction  of  Cii'culates 123 

Addition  of  Circulates 124 

Subtraction  of  Circulates .  .  125 

Multiplication  of  Circulates 126 

Division  of  Circulates  .  127 


CONTENTS. 


FAOR. 

COMPOUND  NUMBERS 12.3 

Long  or  Linear  Measure — Mariners'  Measure 120 

Surveyors'  and  Engineers'  Measure HjO 

Cloth  Measurer 130 

Square  and  Surface  Measure 131 

Land  Measure         132 

Cubic  or  Solid  Measure 133 

Troy  or  Miut  Weight 134 

Diamond  Weight 135 

Apothecaries  Weight        loo 

Avoirdupois  or  Commercial  Weight 130 

Comparison  of  Weights 136 

Wine  Measure 137 

Ale  and  Beer  Measure 137 

Dry  Measure 138 

Comparison  of  Measures 138 

Apothecaries'  Fluid  Measure         139 

Time 139 

Circular  or  Angular  Measure 142 

Comparison  of  Time  and  Longitude        142 

Federal  or  United  States  Money        143 

English  or  Sterling  Money         145 

State  Currencies 146 

French  Weights,  Measures,  and  Money 147 

Foreign  Weights  and  Measures 148 

Reduction  of  Compound  Numbers, 151 

Addition  of  Compound  Numbers        166 

Subtraction  of  Compound  Numbers 168 

Multiplication  of  Compound  Numbers 171 

Division  of  Compound  Numbers 175 

ALIQUOT  PARTS       181 

Exercises  in  State  Currencies        181 

Miscellaneous  Examples  in  Compound  Numbers 180 

RATIO H7 

PROPORTION 1- 

Simple  Proportion 10 1 

Compound  Proportion  * I!'1.) 

Rule  of  Cause  and  Effect 200 

PERCENTAGE 202 

To  Find  any  given  Per  Cent,  of  a  Number 20:i 

To  Find  the  Rate  Per  Cent,  one  number  is  of  another  .     .          .  200 
To  Find  a  Number,  when  a  certain  per  cent,  oi  it  is  known       .  208 
A.  Number  being  given,  which  is  a  certain  per  cent, 
•noro  or  less  than  another,  to  find  that  other 210 


CONTENTS.  Vli 

* 

PAOl. 

APPLICATIONS  OP  PERCENTAGE 212 

Gain  and  Loss 213 

Commission  and  Brokerage 219 

Stocks  and  Dividends 226 

Par,  Discount,  and  Premium 227 

Insurance 232 

Taxes 236 

Duties  or  Customs 239 

INTEREST 242 

Simple  Interest . 244 

Practical  Rules  for  computing  Simple  Interest 247 

Present  Worth  and  Discount 263 

BANKING 256 

Promissory  Notes 257 

To  Find  the  day  a  Note  is  legally  due 268 

DISCOUNTING  NOTES 258 

To  Find  the  Proceeds  of  a  Note 260 

To  Find  the  rate  of  Interest,  when  a  Note  is  Discounted  .  263 
To  Find  the  Face  of  any  Note,  when  the  proceeds,  time, 

and  rate  of  Discount,  are  given 264 

To  find  the  rate  of  Discount,  corresponding  to  a  given  rate 

of  Interest 266 

Rules  for  Partial  Payments 266 

EXCHANGE 271 

Table  of  Foreign  Coins  and  moneys  of  account 274 

Home  or  Inland  Exchange 275 

Foreign  Exchange 276 

Arbitration  of  Exchange 278 

Chain  Rule 279 

ACCOUNTS  CURRENT 282 

To  settle  an  Account  Current  whose  items  draw  Interest    .  283 

To  settle  a  Merchant's  Account  Current 286 

STORAGE  ACCOUNTS 288 

EQUATION  OP  PAYMENTS 289 

COMPOUND  INTEREST 295 

To  find  the  Comp.  Int.,  principal,  rate  and  time  given  .  .  296 
To  find  the  Rate,  when  the  principal,  time,  and  compound 

interest  or  amount  is  given 300 

T.o  find  the  Principal,  time,  rate,  and  interest  given  .  .  .  801 

/  The  same,  time,  rate,  and  compound  amount  given  .  .  302 
To  find  the  Time,  when  the  principal,  rate,  and  compound 

interest  or  amount,  are  given ...  303 


viii  CONTENTS. 


PAQB 

ANNUITIES 306 

Ti  find  the  Initial  value  of  a  Perpetuity 804 

To  find  the  Present  value  of  a  deferred  Perpetuity  .  .  .  305 

To  find  the  Present  value  o.f  an  Annuity  certain  ....  306 

To  find  the  Forborne  or  final  value  of  an  Annuity  .  .  .  .  307 
To  find  an  Annuity,  when  its  present  or  final  value,  rate 

of  Interest,  and  time  to  run,  are  known 309 

To  find  the  time  a  given  annuity  runs,  when  the  rate  of 

Interest  and  present  or  final  value  are  known  ....  309 
To  find  the  rate  of  Interest  of  a  given  annuity,  the  present 

value,  and  time  to  run,  known 310 

Contingent  Annuities 311 

To  find  the  value  of  a  Life-Estate  or  Widow's  dower  .  .  .  311 
To  find  how  large  a  Life  Annuity  can  be  purchased  for  a 

given  sum 313 

To  find  the  value  of  the  reversion  of  a  Life  Annuity  .  .  .  313 
To  find  the  Single  and  Annual  Premiums  necessary  to 

secure  a  given  sum  at  the  death  of  the  person  Insured  .  313 

To  find  the  value  of  a  Policy  of  Life-Insurance 314 

PROPORTIONAL  PARTS 315 

Partnership 317 

Partnership  with  Time 319 

Bankruptcy 321 

General  average 322 

Rate  Bills  for  Schools 324 

Alligation 325 

INVOLUTION 830 

EVOLUTION 332 

Square  root 332 

Cube  Root 338 

Extraction  of  any  Root . 342 

SERIES 344 

Arithmetical  Series 845 

Geometrical  Series .    .     .    ,    .     .  346 

PERMUTATIONS,  COMBINATIONS 848 

SYSTEMS  OP  NOTATION 349 

DUODECIMALS '  ...  350 

MENSURATION 351 

Plasterers',  Painters',  and  Pavers'  Work 35'J 

Mensuration  of  Solids 354 

Masons'  and  Bricklayers'  Work 357 

Guaging 358 

Tunnage  of  Vessels S'">3 

MECHANICAL  POWERS see 


ARITHMETIC. 


I.  INTRODUCTION. 

ARTICLE  1.  QUANTITY  is  any  thing  which  can  be  in- 
creased or  diminished.  Thus,  numbers,  lines,  space,  motion, 
time,  and  weight,  are  quantities. 

ART.  2.   MATHEMATICS  is  the  science  of  quantity. 

ART.  3.  The  fundamental  branches  of  Mathematics  are 
Arithmetic,  Algebra,  and  Geometry. 

ART.  4.   ARITHMETIC  is  the  science  of  numbers. 

ART.  5.   A  PROBLEM  is  a  question  proposed  for  solution. 

ART.  6.   A  THEOREM  is  a  truth  to  be  proved. 

ART.  7.  Problems  and  Theorems  are  both  called  Pro- 
positions. 

ART.  8.  A  COROLLARY  is  a  truth  deduced  from  a  pre- 
ceding proposition. 

ART.  9.  A  DEMONSTRATION  is  a  process  of  reasoning 
by  which  a  proposition  is  shown  to  be  true. 

ART.  10.  A  DIRECT  DEMONSTRATION  is  one  which  com- 
mences with  known  truths,  and  by  a  chain  of  reasoning 
establishes  the  proposition  to  be  proved. 

REVIEW.  — 1.  What  is  quantity  ?  Give  examples.  2.  What  is  Mathe- 
matics? 3.  What  are  its  fundamental  branches?  4.  Define  Arithmetic. 
6.  What  is  a  problem?  6.  A  theorem  ?  7.  What  common  name  is  applied 
to  both?  3  What  is  a  corollary?  9.  A  demonstration?  10.  A  direct 
demonstration  ? 

(9) 


10  RAY'S   HIGHER   ARITHMETIC. 

ART.  11.  An  INDIRECT  DEMONSTRATION  is  one  which 
assumes  the  proposition  to  bo  fake,  and  then  proves  that 
some  absurdity  will  necessarily  follow.  This  is  sometimes 
palled  Reductio  ad  absurdum. 

ART.  12.  An  AXIOM  is  a  self-evident  truth ;  that  is,  a 
proposition  so  evident  that  it  can  not  be  made  plainer  by 
any  demonstration.  The  following  are  among  the  most 
important  axioms. 

1.  If  the  same  or  equal  quantities  be  added  to  equals, 
the  sums  will  be  equal. 

2.  If  the  same  or  equal  quantities  be  subtracted  from 
equals,  the  remainders  will  be  equal. 

3.  If  the  same  or  equal  quantities  be  multiplied  by  the 
game  number,  the  products  will  be  equal. 

4.  If  the   same  or  equal  quantities  be  divided  by  the 
same  number,  the  quotients  will  be  equal. 

5.  Generally,  if  the  same  identical   operation   be   per- 
formed on  two  equal  quantities,  the  results  will  be  equal. 

MATHEMATICAL  SIGNS. 

ART.  13.  For  brevity,  characters,  called  signs,  are  used 
in  Mathematics.  Those  most  used  in  Arithmetic,  are 


+,     — ,     X,     -=-,     =,     (  )or 


ART.  14.  The  sign  -}-,  read  plus,  is  the  sign  of  Addi- 
tion; it  shows  that  the  numbers  batween  which  it  is  placed 
are  to  be  added  together.  Thus,  3  +  5  equals  8. 

ART.  15.  The  sign  — ,  read  minus,  is  the  sign  of  Sub- 
traction; the  number  which  follows  it  is  to  be  subtracted 
from  that  which  precedes  it.  Thus,  7  —  4  equals  3. 

Plus  and  minus  are  Latin  words,  signifying  more  and  less. 

ART.  18.  The  sign  X,  read  times,  is  the  sign  of  Multi- 
plication. The  numbers  between  which  it  is  placed  are  to 
Be  multiplied  together.  Thus,  4x5  equals  20. 

ART.  17.  The  sign  -7-,  read  divided  by,  is  the  sign  of 
Division.  The  number  which  precedes  it  is  to  be  divided 
by  that  which  follows  it.  Thus,  20  -j-  4  equals  5. 

REVIEW. — 11.  An  indirect  demonstration?  12.  What  is  an  axiom  1 
Name  the  most  important  axioms.  13-19.  Describe  the  signs  most  fre- 
tnently  need  in  Arithmetic,  and  give  examples  of  their  use. 


NUMERATION   AND  NOTATION. 


ART.  18.  The  sign  =,  read  equate,  or  is  equal  to,  is  the 
sign  of  Equality;  the  quantities  between  which  it  is  placed 
are  equal  to  each  other.  Thus,  5  -f-  3  =  9  —  1. 

ART.  19.  A  parenthesis  (  ),  or  vinculum  -  :,  shows 
that  two  or  more  numbers  are  to  be  considered  as  •  one. 
Thus,  (7  +  4)  X  3  =  33;  or  8^5x4=19  +  5-^2. 

ARITHMETICAL   DEFINITIONS. 
ART.  20.    A  unit  is  a  single  thing;  a  cent,  a  hat,   &o. 

ART.  21.  A  number  is  a  unit,  or  a  collection  of  units 
classed  under  the  same  name,  and  answers  to  the  question, 
Sow  many? 

The  unit  of  a  number  is  one  of  the  things  it  expresses: 
thus,  in  Jive  cents,  one  cent  is  the  unit;  in  ten  apples,  one 
apple  is  the  unit. 

Units  are  sometimes  only  relative  in  their  character; 
thus,  one  foot  is  a  unit  in  regard  to  feet,  but  it  is  only  a 
part  of  a  unit  in  regard  to  yards.  One  din?  a  is  a  unit,  that 
is,  one  in  regard  to  dimes,  but  it  is  ten  in  regard  to  cents. 

ART.  22.  Numbers  are  either  abstract  or  concrete.  An 
abstract  number  is  one  in  which  the  kind  of  unit  is  not 
designated,  as  one,  two,  three,  &c.  A  concrete  number  is 
one  in  which  the  kind  of  unit  is  designated,  as  one  cent, 
two  apples,  ten  bushels,  &c.  Concrete  numbers  are  fre- 
quently called  denominate  numbers. 

ART.  23.  Arithmetic  is  founded  on  NOTATION,  and  its 
operations  are  carried  on  by  means  of  ADDITION,  SUBTRAC- 
TION, MULTIPLICATION,  and  DIVISION.  These  are  termed 
the  fundamental  rules  of  Arithmetic. 


II.    NUMERATION  AND   NOTATION. 

ART.  24.    NUMERATION  is  the  art  of  naming  numbers. 
NOTATION  is  the  art  of  representing  numbers  by  char- 
acters called  figures  or  digits. 

REVIEW.— 20.  What  is  a  unit?  21.  What  a  number?  Show  th« 
relative  character  of  some  units.  22.  What  arc  abstract  numbers  ?  Con- 
crete? Give  examples.  23.  On  what  is  Arithmetic  founded?  What  art 
its  fundamental  rules  ?  24.  What  is  numeration  ?  What  is  notation  ? 


12  RAY'S   HIGHER   ARITHMETIC. 


ART.  25.  The  first  nine  numbers  are  each  represented 
by  a  single  figure,  thus: 

12345        6        T89 

one.  ,  two.      three,    four.     fire.      iix.     seven,  eight,  nine. 

All  other  numbers  are  represented  by  combinations  of 
these  and  another  figure,  0,  called  zero,  naught,  or  cipher. 

REMARK  . — The  cipher,  0,  is  used  to  indicate  no  value.  The  other 
flgures  are  called  significant  figures,  because  they  indicate  some  value. 

* 

ART.  26.  The  number  next  higher  than  9  is  named 
TEN,  and  is  written  with  two  figures  thus,  10:  in  which 
the  cipher,  0,  merely  serves  to  show  that  the  unit,  1,  on 
its  left  is  different  from  the  unit,  1,  standing  alone,  which 
represents  a  single  thing,  while  this  represents  a  single 
group  of  ten  things. 

The  numbers  succeeding  Ten  are  written  and  named  as 
follows : 

11  12  13  14  15  16 

eleven.         twelve.      thirteen,     fourteen,      fifteen.        sixteen. 

17          18          19 

seventeen,   eighteen,     nineteen. 

In  each  of  which,  the  1  on  the  left,  represents  a  group  of 
ten  things,  while  the  figure  on  the  right,  expresses  the 
units  or  single  things  additional,  required  to  make  up  the 
number. 

REMAKE. — The  words  eleven  and  twelve  are  supposed  to  be  de- 
rived from  the  Saxon,  meaning  one  left  after  ten,  and  two  left  after  ten. 
The  words  thirteen,  fourteen,  &c.,  are  contractions  of  three  and  ten, 
four  and  ten,  &c. 

The  next  number  to  nineteen,  (nine  and  ten),  is  ten  and 
ten,  or  two  groups  of  ten,  written  20,  and  called  twenty. 

The  next  are  twenty-one,  21;  twenty-two,  22;  &c.,  up 
to  tfiree  tens,  or  thirty,  30;  forty,  40;  fifty,  50;  sixty,  60; 
eventy,  *70;  eighty,  80;  ninety,  90. 

RE vi E  w. — 25.  How  are  the  first  nine  numbers  represented?  How  are 
numbers  above  nine  represented?  Rem.  For  what  is  the  cipher  used? 
What  are  the  other  figures  called?  26.  How  is  Ten  written?  What 
does  the  cipher  show?  What  does  the  1  represent?  Write  the  next  nine 
bombers.  Explain  their  names.  Show  what  the  figures  denote  in  each. 


NUMERATION   AND   NOTATION.  13 

The  highest  number  that  can  be  written  with  two  figures 
is  99,  called  ninety-nine;  that  is,  nine  tens  and  nine  units. 

The  next  higher  number  is  9  tens  and  ten,  or  ten  tens, 
which  is  called  a  hundred,  and  written  with  three  figures, 
100;  in  which  the  two  ciphers  merely  show  that  the  unit 
on  their  left  is  neither  a  single  thing,  1,  nor  a  group  of 
ten  things,  10,  but  a  group  of  ten  tens,  being  a  unit  of  a 
higher  grade  than  either  of  those  already  known.  ' 

In  like  manner,  200,  300,  &c.,  express  two  hundreds, 
three  hundreds,  and  so  on  up  to  ten  hundreds,  called  a 
thousand,  and  written  with  four  figures,  1000,  being  a 
unit  of  a  still  higher  order. 

ART.  27.  From  what  has  been  said,  it  is  clear  that  a 
figure  in  the  1st  place,  with  no  others  to  the  right  of  it, 
expresses  units  or  single  things;  but  standing  on  the  left 
of  another  figure,  that  is,  in  the  2d  place,  expresses  groups 
of  tens;  and  standing  at  the  left  of  two  figures,  or  in  the 
3d  place,  expresses  tent  of  tens,  or  hundreds;  and  in  th.e 
4th  place,  expresses  tens  of  hundreds  or  thousands.  Hence, 
counting  from  the  right  hand, 

The  order  of  Units           is  in  the  1st  place  .  1 

The  order  of  Tens            is  in  the  2d   place  .  10 

The  order  of  Hundreds  is  in  the  3d   place  .  100 

The  order  of  Thousands  is  in  the  4th  place  .  1000 

By  this  arrangement,  the  same  figure  ha»  different  values 
according  to  the  place  in  which  it  stands.  Thus,  3  in  the 
first  place  is  3  units;  in  the  second  place  3  tens,  or  thirty; 
in  the  third  place  3  hundreds;  and  so  on. 

ART.  28.  The  word  UNITS  may  be  used  in  naming  all 
the  orders  as  follows: 

Simple  units  are  called  .     .     .  Unit*  of  the  1st  order. 

Tens Units  of  the  2d  order. 

Hundreds Units  of  the  3d  order. 

Thousands Units  of  the  4th  order. 

&c.  &o. 

RKTIHW.— 26.  What  are  20,  30,  40  ?  What  is  the  highest  number  of  two 
figures  ?  What  Is  tho  next  called  ?  How  is  it  written  ?  Explain  its  figures. 
What  is  a  Thousand?  How  written?  27.  What  does  a  figure  in  the  1st 
place  express?  a  figure  in  the  2d  place?  in  the  3d  place?  in  the  4tb 
place  ?  How  doe*  tho  value  of  the  same  figure  vary  ? 


i  i 


BAY'S   HIGH  Ell    ARITHMETIC. 


NOTE. — Wlicu  units  are  named  without  reference  to  any  par- 
ticular order,  units  of  the  first  order  are  always  intended. 

ART.  29.  The  preceding  articles  show  the  method  of 
expressing  numbers  less  than  one  thousand.  For  exam- 
ple, in  the  number  four  hundred  and  twenty-jive,  there  are 
4  hundreds,  2  tens  or  twenty,  and  5  units;  the  number 
is  therefore  written,  425. 

In  the  number  three  hundred  and  nine,  there  arc  3  hun- 
dreds, 0  tens,  and  9  units ;  or  3  units  of  the  third,  and 
none  of  the  second,  and  9  of  the  first  order ;  hence,  tho 
number  is  represented  thus,  309. 

ART.   30.       SUMMARY   OF   PRINCIPLES. 

1.  All  numbers  are  represented  by  the  nine  digits  and  zero 

2.  Zero  has  no  value;  its  use  is  to  Jill  vacant  pin <:<•*. 

3.  The  base  of  our  system  of  notation  is  ti-n  ;   ten   units  of 
any  order  making  one  unit  of  the  order  next  higher. 

4.  The  same  figure  has   different   values   according  to  the 
place  it  occupies. 


2d.      1st. 


ART. 

31.               TABLE 

OF   ORDERS. 

9th. 

8th.     7th.     6th. 

5th.      4th.       3( 

• 

BB 

• 

• 

• 

a 

• 

a 

. 

«4 

• 

_o 

, 

<n 

m 

S 

00 

a 

0 

H 

a 

05 

CM 

CM 

0 

O 

uj 

O 

no 

>*H 

•B 

_ 

F^              no 

rr^                       «. 

§ 

no  .pj 


-0 

a 

3 
W 


a 
<u 

EH 


a 

cj 

in 

S 
O 

-= 

H 


T5 

a 

PJ 


3 


AET.  32.  For  convenience  in  reading  and  writing  num- 
bers, periods  of  three  orders  each,  are  used.  The  first  3 
orders,  UNITS,  TENS,  and  HUNDREDS,  constitute  the  first 
or  UNIT  PERIOD.  The  second  3  orders  form  the  second 
or  THOUSAND  PERIOD;  the  third  3  orders,  the  third  or 

MILLION  PERIOD. 


REVIEW. — 29.  Write  the  number  four  hundred  and  twenty-fire.    Write 
'he  number  threo  hundred  and  niuo. 


AND   NUMERATION  15 


ART.  33.    List  of  the  Periods,  according  to  the  commcn 
or  French  method  of  Numeration. 


First  Period  .  Units. 

Second  .    .    .  Thousands. 

Third     .    .    .  Millions. 

Fourth  .     .    .  Billions. 

Fifth  .  Trillions. 


Sixth  Period  .  Quadrillions. 
Seventh  .    .    .  Quintillions. 
Eighth   .    .    .  Sextillions. 
Ninth     .    .    .  Septillions. 
Tenth  .  Octillions. 


The  next  twelve  periods  are,  Nonillions,  Decillions,  Un- 
decillions,  Duodecillions,  Tredecillions,  Quatuordecillions, 
Quindecillions,  Sexdecillions,   Septendecillions,  Octodecil 
lions,  Novendecillions,  Vigintillions. 

ART.  34.    Division  of  the  orders  into  periods. 


o  ^~"j  o  ^ 

£  m 


•a 

1 

§  g  'S 

W  H  f= 

654 

5th  Period. 

n 

"3           oo 

s  g  'a 

B  H  & 

321 

4th  Period. 

••o 
E 

||| 

987 

3d  Period. 

13 
p 
E 

1    gl 

tC      &^      t^ 

654 

2d  Period. 

•3 
« 

b 
"•^              CO 

§  i's 

B  H  & 

321 

1st  Period. 

ART.  35.      RULE   FOR  NUMERATION. 

Begin  at  the  right,  and  point  the  number  into  periods  of  tliret 
figures  each.  Commence  at  the  left,  and  read  in  succession  each 
period  with  its  name. 

REMARK. —  Numbers  may  also  be  read  by  merely  naming  each 
figure  with  the  name  of  the  place  in  which  it  stands.  This  method, 
however,  is  rarely  used  except  in  teaching  beginners.  Thus,  the 
numbers  expressed  by  the  figures  205,  may  be  read  two  hundred  and 
five,  or  two  hundreds  no  tens  and  five  units. 

Express  in  words,  the  number  which  is  represented  by 
608921045.  The  number,  a?  divided  into  period?,  is  608* 
J)21'045 ;  and  is  read  six  uundred  and  eight  millions 
nine  hundred  and  twenty  ne  thousand  and  forty-five. 


REVIEW. — 30.  What  are  '  o  principles  of  Notation  and  Numeration T 
81.  Repeat  the  Table  of  O- ier8.  32.  What  are  periods  and  their  uBel 
33.  Name  the  first  ten.  3-4  tlive  an  example  of  tho  use  of  periods  in  a  par- 
ticular case.  35.  What  is  the  rule  for  Numeration  ?  What  other  method 
may  ho  used? 


(6  RAY'S   HIGHEll   ARITHMETIC. 


EXAMPLES   IN   NUMERATION. 

7     12345    1375482     29347283 

40     68380    6030564     37053495 

85     94025    7004025     45004024 

503     70500.   8025607     50340726 

278    165247    9030040     60025709 

1345    350304    6002007    343827544 

2450    204026    4300201    904207080 

3708    500050    8603004    700200408 

4053    808080    2030405    502003070 

7009    730003    6005010    830070320 

832045682327825000000321 

8007006005004003002001000000 

60030020090080070050060030079 

504030209102800703240703250207 

ART.  36.    ROLE  FOR  NOTATION. 

Write,  first  the  number  of  the  highest  period,  then,  of  (he  other 
periods  in  their  proper  succession,  filling  vacant  places  with 
ciphers. 

Express  in  figures  the  number  four  millions  twenty 
thousand  three  hundred  and  seven.  Ans.  4020307. 

Write  4  in  millions  period ;  place  a  dot  after  it  to  sepa- 
rate it  from  the  next  period :  then,  write  20  in  thovtandt 
period ;  place  another  dot :  then  write  307  in  units  period. 
This  gives  4'  20*307.  As  there  are  but  two  places  in 
the  thousands  period,  a  cipher  must  be  put  before  20  to 
complete  its  orders,  and  the  number  correctly  written,  is 
4020307. 

NOTE. —  Every  period,  except  the  highest,  must  have  its  three 
figures;  and  if  any  period  is  not  mentioned  in  the  given  number, 
supply  its  place  by  three  ciphers. 

PROOF. — Apply  to  the  number,  as  written,  the  rule  for 
numeration,  and  see  if  it  agrees  with  the  number  given. 

EXAMPLES   IN    IS     TATION. 


1.  Seventy-five. 

2.  Ninety. 

3.  One  hundred    and    thirty- 
four. 


4.  i  <  o    hundred    and    forty- 
four 

5.  Two  hundred  and  forty. 

6.  Two  hundred  and  four. 


NUMERATION    AND   NOTATION. 


/ .   Three  hundred  and  seventy. 

8.  Six  hundred  and  ten. 

9.  Eight  hundred  and  one. 

10.  One  thousand  two  hundred 
and  thirty-four. 

11.  Three  thousand  four  hun- 
dred and  ninety. 

12.  Seven  thousand  and  twenty- 
five. 

13.  Nina  thousand  and  seven. 

14.  Forty   thousand  five    hun- 
dred and  sixty-three. 

15.  Seventy  thousand  and  twen- 
ty. 

16.  Eighty-four  thousand    and 
'  two. 

17.  Ninety  thousand  and  nine. 

18.  Sixty    thousand    six    hun- 
dred. 

19.  One  hundred  and  sixty-four 
thousand  three  hundred  and 
ninety-four. 

20.  Two    hundred    and    seven 
thousand  four  hundred  and 
one. 

21.  Five  hundred  thousand  and 
thirty. 

22.  Six  hundred  and  forty  thou- 
sand and  forty. 

23.  Eight    hundred     and    two 
thousand  two  hundred. 

24.  Nine     hundred      thousand 
nine  hundred. 

25.  Seven     hundred    thousand 
and  seven. 

26.  One  million  four  hundred 
and    twenty-one    thousand 
six    hundred    and    eighty- 
five. 

27.  Six  millions  sixty  thousand 
and  sixty. 

28.  Nine  millions    nine    thou- 
sand and  nine. 

29.  Seven  millions  and  seventy- 


30.  Twenty-three  millions    sia 
hundred  and  forty-five  thou- 
sand nine  hundred  and  sev- 
enty-one. 

31.  Ten  millions  one  hundred 
thousand  and  ten. 

32.  Ninety  millions  ninety  thou- 
sand and  ninety. 

33.  Eighty-eight  millions    sev- 
enty thousand  and  five. 

34.  Sixty   millions  seven   hun- 
dred and  five  thousand. 

35.  One  hundred  millions. 

36.  Two  hundred  and  fifty  mil- 
lions   three    hundred    and 
three  thousand  and  twenty- 
six. 

37.  Eight  hundred   and   seven 
millions  forty  thousand  and 
thirty-one. 

38.  Seven  hundred  millions  sev- 
enty thousand  and  seven. 

39.  Two   billions    and    twenty 
millions. 

40.  Thirty  trillions  thirty  mil- 
lions thirty   thousand   and 
thirty. 

41.  Nineteen  quadrillions  twen- 
ty  trillions  and  five  hun- 
dred billions. 

42.  Ten  quadrillions  four  hun- 
dred   and    three    trillions 
ninety     billions     and     six 
hundred  millions. 

43.  Eighty  octillions  sixty  sex- 
tillions  three  hundred  and 
twenty-five  quintillions  and 
thirty-three  billions. 

44.  Nine     hundred     decilliona 
seventy  nonillions   six    oc- 
tillions forty  septillions  fif- 
ty   quadrillions    two    hun- 
dred and  four  trillions  ten 
millions  forty  thousand  and 
sixty. 


15  R  v  i  K  w.— 36.  What  is  the  rule  for  notation  ?     What  is  the  proof  ? 
2 


18  RAY'S   HIGHER   ARITHMETIC. 


ENGLISH   METHOD    OF   NUMERATION. 

ART.  37.  Although  now  but  little  used,  the  learnei 
should  understand  this  system  of  Numeration.  The  first 
six  orders  have  the  same  names  as  in  the  French  method, 
and  these  constitute  the  period  of  Units.  The  next  01 
Millions  period,  consists  of  six  orders.  Each  succeeding 
period  consists  of  six  orders,  and  their  names  are  Billions, 
Trillions,  Quadrillions,  Quintillions,  and  so  on.  The  fol- 
lowing table  illustrates  this  method: 


a 
«« .2 


oo  EH  T5  _?o  J£  E~<  "O  J2  oo  f-^  f-j   co  _2  H  T3  JS 

g3*v  Q1""^  ^          p  ^-*                                   fl   X 

s-oS—  t-oS*'  ^oS^  uo?s- 

•W             2t3             00  T3             2*0             00  rrt             S^             00  T3             "•   " "             00 

^MS^oo-i  2oo32on-t>  Soo3rtOQ-S  XooSSm" 

lulll  IlilJg  lljlls 


432109  876543  210987  654321 

By  this  system  the  first  twelve  figures  would  be  read, 
two  hundred  and  ten  thousand  nine  hundred  and  eighty- 
seven  millions,  six  hundred  and  fifty-four  thousand  three 
hundred  and  twenty-one.  By  the  French  method  they 
would  be  read,  two  hundred  and  ten  billions,  nine  hun- 
dred and  eighty-seven  millions,  six  hundred  and  fifty-four 
thousand,  three  hundred  and  twenty-one. 

ROMAN   NOTATION. 

ART.  38'.  In  the  Roman  Notation,  numbers  are  repre- 
sented by  letters.  The  letter  I  represents  one;  V,  five;  X, 
ten;  L,  fifty,  C,  one  hundred;  D,  five  hundred;  and  M,  one 
thousand.  The  other  numbers  are  represented  according 
to  the  following  principles: 

1st.  Every  time  a  letter  is  repeated,  its  value  is  repeated. 
Thus,  II  denotes  two;  XX  denotes  twenty. 

REVIEW. — 37.  Explain  the  English  method  of  Numeration.  Give  an 
example  of  ite  application.  38.  How  are  numbers  represented  in  th« 
Roman  Notation  ?  What  numbers  are  repregentod  by  I,  V,  X,  C,  D,  M  ? 


ADDITION. 


2d.  Where  a  letter  of  less  value  is  placed  before  one  of 
greater  value,  the  less  is  taken  from  the  greater.  If  placed 
after  it,  the  less  is  added  to  the  greater.  Thus,  IV  denotes 
four,  while  VI  denotes  six ;  IX  denotes  nine,  while  XI 
denotes  eleven. 


3d.  A 

thousand 

bar,  —  ,  placed  over  a  letter  increases  its  valu*  a 
times.     Thus,  V  denotes  five  thousand;  M  denotes 

one  million. 

ROMAN   TABLE. 

I        . 

.     .   One.                i  XXX  . 

.    Thirty. 

II      . 

.     .   Two. 

XL      . 

.    Forty. 

Ill    . 

.     .   Three. 

L 

.    Fifty. 

IV    . 

.     .    Four. 

LX      . 

.   Sixty. 

V      . 

.     .   Five. 

xc    . 

.    Ninety. 

VI    . 

.     .   Six. 

c 

.    One  hundred. 

IX    . 

.     .   Nine. 

cccc  . 

.    Four  hundred. 

X      . 

.     .    Ten. 

D 

.    Five  hundred. 

XI    . 

.    Eleven. 

DC      . 

.   Six  hundred. 

XIV 

.     .    Fourteen. 

DCC    . 

.    Seven  hundred. 

XV  . 

.     .    Fifteen. 

DCCC  . 

.   Eight  hundred. 

XVI 

.    Sixteen. 

DCCCC 

.    Nine  hundred. 

XIX 

.    Nineteen. 

M 

.    One  thousand. 

XX  . 

.     .    Twenty. 

MM     . 

.    Two  thousand. 

XXI 

.     .   Twenty-one.      MDCCCLVI      1856. 

III.    ADDITION. 

ART.  39.  ADDITION  is  the  process  of  collecting  two  or 
more  numbers  into  one  sum. 

Sum  or  Amount  is  the  result  obtained  by  Addition. 

ART.  40.  Since  a  number  is  a  collection  of  units  of  the 
tame  kind,  two  or  more  numbers  can  be  united  into  one  turn, 
only  when  their  units  are  of  the  same  kind.  Two  apples  and 
3  apples  are  5  apples;  but  2  apples  and  3  peaches  can  not 
be  united  into  one  number,  either  of  apples  or  of  peaches. 


REVIEW. — 38.  What  ia  the  effect  of  repeating  a  letter?  Of  placing  • 
letter  of  less  value  before  another  of  greater  value  ?  Of  placing  one  of  loss 
value  after  one  of  greater  value  ?  Of  placing  a  bar  over  a  letter?  39.  What 
ta  addition  ?  Bum  or  amount  T 


20  RAY'S  HIGHER  ARITHMETIC. 

Nevertheless,  numbers  of  different  names  may  be  added 
together,  if  they  can  be  brought  under  a  common  denomi- 
nation. Thus,  2  men  and  3  women  are  5  persons.  Two 
horses,  3  sheep,  and  4  cows,  are  9  animals. 

RULE  FOR  ADDING  SIMPLE  NUMBERS. 

1.  Write  the  numbers  to  be  added,  so  that  figures  of  the  same 
order  may  stand  in  column,  units  under  units,  tens  under  tens, 
hundreds  under  hundreds,  &c. 

1.  Begin  at  the  right,  and  add  each  column  separately,  placing 
the  units  of  each  sum  under  the  column  added,  and  carrying  the 
tens  to  the  next  column.     At  the  last  column,  set  down  the  whole 
amount. 

What  is  the  sum  of  639,  82,  and  543? 

SOLUTION. — Writing  the  numbers  as  in  the  margin,         639 
say,  3  and  2  are  5,  and  9  are  14  units,  which  are  1  ten  8  2 

and  4  units.     Write  the  4  units  beneath,  and  carry  the         543 
I  ten  to  the  next  column.    Then  1  and  4  are  5,  and  8  are     i  O  CJ. 
13,  and  3  are  16  tens,  which  are  6  tens   to  be  written 
beneath,  and  1  hundred  to  be  carried  to  the  next  column.     Lastly, 
I  and  5  are  6,  and  6  are  12  hundreds,  which  is  set  beneath,  there 
being  no  other  columns  to  carry  to  or  add. 

DEMONSTRATION. — 1.  Figures  of  the  same  order  are 
written  under  each  other  for  convenience,  since  none  but 
units  of  the  same  name  can  be  added.  (Art.  40.) 

2.  Commence  at  the  right  to  add,  so  that  when  the  sum 
of  any  column  is  greater  than  nine,  the  tens  may  be  car- 
ried to  the  next  column,  and,  thereby,  units  of  the  same 
name  added  together. 

3.  Carry  one  for  every  ten,  since  ten  units  of  each  order 
make  one  unit  of  the  order  next  higher.     (Art.  30.) 

METHODS   OF   PROOF 

ART.  41.  1.  Add  the  figures  downward  instead  of  up- 
ward;  or 


R  E  vi  K  w. — 40.  What  is  necessary  in  order  that  numbers  may  be  added 
into  one  sum  ?  Why  ?  What  is  the  rule  for  adding  simple  numbers  ? 
Explain  the  example,  and  give  the  reasons  for  the  rule.  41.  What  is  the 
Erst  method  of  proof? 


ADDITION.  21 


2.  Separate  the  numbers  into  two  or   more    divisions; 
find  the  sum  of  the  numbers  in  each  division,  and  then 
add  the  several  sums  together;  or 

3.  Commence  at  the  left;  add  each  column  separately; 
place  each  sum  under  that  previously  obtained,  but  ex- 
tending one  figure  further  to  the  right,  and  then  add  them 
together. 

In  all  these  methods  the  result  should  be  the  same  as 
when  the  numbers  are  added  upward. 

NOTE. — For  proof  by  casting  out  the  9's,  see  page  68. 


EXAMPLES  FOR  PRACTICE. 
Find  the  sum, 

1.  Of  76767;  7654;  50121;  775.     Ant.  135317. 

2.  Of  97674;  686;  7676;  9017.       An*.  115053. 

3.  Of  971;  7430;  97476;  76734.     AM.  182611. 

4.  Of  four  hundred  and  three;   5025;  sixty  thousand 
and   seven;    eighty-seven    thousand;    two    thousand    and 
ninety,  and  100.  Ana.  154625. 

5.  Of  999;  3400;  73;  47;  452;  11000;  193;  97; 
9903;  42,  and  5100.  AM.  31306. 

6.  Of  20050;    three   hundred  and  seventy  thousand 
two  hundred;  four  millions  and  five;  two  millions  ninety 
thousand  seven  hundred  and  eighty ;  one  hundred  thou- 
sand and  seventy;  98002;  seven  millions   five  thousand 
and  one;   and  70070.  Am.  13754178. 

7.  Of  609505;  90070;  90300420;  9890655;  789; 
37599;    19962401;    5278;    2109350;    41236;    722; 
8764;  29753,  and  370247.  Am.  123456789. 

8.  Of  two  hundred  thousand  two  hundred;  three  hun- 
dred millions  six  thousand  and   thirty;    seventy  millions 
seventy   thousand  and   seventy;   nine  hundred   and    four 
millions  nine  thousand  and  forty;  eighty  thousand ;  ninety 
millions  nine  thousand;  six  hundred  thousand  and  sixty; 
five  thousand  seven  hundred;  four  millions  twenty  thou- 
sand and  twenty.  Ans.  1369000120, 

RBVIKW. — What  is  the  second  method  of  proof?     The  third  ? 


2<<>  RAY'S   HIGHER    ARITHMETIC. 


In  each  of  the  7  following  examples,  find  the  sum  of 
the  numbers  from  A  to  B,  including  these  numbers. 

A.  B. 

9.         119    ...       131     ...  Ans.         1625. 

10.  987    ...     1001     .     .     .  Ans.       14910. 

11.  3267    .    .    .     3281     .    .    .  Ans.      49110. 
12        4197    .    .     .     4211     .     .     .  Ans.       63060. 

13.  5397    .    .    .     5416    .    .    .  Ans.    108130. 

14.  7815    .     .     .     7831     .     .     .  Am.     132991. 

15.  31989    .     .     .  32028     .     .     .  Ans.  1280340. 

16.  Paid  for  coffee,  $375;    for  tea,   $280;    for  suga,, 
$564;    for   molasses,  $119;    and   for  spices,   $75:    what 
did  the  whole  amount  to?  Ans.  $1413. 

17.  Bought  three  pieces  of  cloth:    the  first   cost  $87; 
the  second,  $25  more  than  the  first;  and  the  third,  $47 
more  than  the  second:    what  did  all  cost?       Ans.  $358. 

.  18.  Bought  three  bales  of  cotton.  The  first  cost  $325; 
the  second,  $16  more  than  the  first;  and  the  third,  as 
much  as  both  the  others:  what  did  the  three  bales  cost? 

Ans.  $1332. 

19.  A  has  $75;  B  has  $19  more  than  A;  C  has  as 
much  as  A  and  B.  and  $23  more;  and  D  has  as  much  as 
A,  B,  and  C  together:  what  sum  do  they  all  possess? 

Ans.  $722. 


IV.  SUBTRACTION. 

ART.  42.  SUBTRACTION  is  the  process  of  finding  the 
difference  between  two  numbers. 

The  larger  number  is  the  Minuend;  the  less,  the  Sub- 
trahend; the  number  left,  the  Difference  or  Remainder. 

Minuend  means  to  bo  diminished;  subtrahend,  to  be. 
subtracted. 

ART.  43.  Subtraction  is  the  reverse  of  Addition,  and 
since  none  but  numbers  of  the  same  kind  can  be  added 

REVIEW.— 42.  What  is  Subtraction?  Minuend?  Subtrahend?  RP 
maindor?  What  does  minuend  mean  ?  What  does  subtrahend  mean  ? 


SUBTRACTION.  93 


together,  (Art.  40),  it  follows,  therefore,  that  a  numbfT 
can  be  subtracted  only  from  another  of  the  same  kind  •. 
2  cents  can  not  be  taken  from  5  apples,  nor  3  cows  from 
8  horses. 

ART.  44.    RULE  FOR   SUBTRACTING  SIMPLE  NUMBERS. 

1.  Write  the  less  number  under  the  greater,  placing  units  under 
units,  tens  under  lens,  &c. 

2.  Begin  at  the  right,  subtract  each  figure  from  the  one  above 
it,  placing  the  remainder  beneath. 

3.  If  any  fgure  exceeds  tlie  one  above  it,  add  ten  to  the  upper, 
subtract  the  lower  from  the  sum,  and  carry  one  to  the  next  lower 
figure. 

From  827  dollars  take  534  dollars. 

SOLUTION. — After  writing  figures  of  the  Barne  ir8' 

order  under  each  other,  say,  4  units  from  7  units          827 
leaves  3  units.     Then,  as  3  tens  can  not  be  taken  534 

from  2  tens,  add  10  tens  to  the  2  tens,  which  makes  298  Rem 

12  tens,  and  3  tens  from  12  tens  leaves  9  tens. 
To  compensate  for  the  10  tens  added  to  the  2  tens,  add  one  hundred 
(10  tens)  to  the  6  hundreds,  and  say,  6  hundreds  from  8  hundreds 
leaves  2  hundreds;  and  the  whole  remainder  is  2  hundreds  9  tens 
and  3  units,  or  293. 

DEMONSTRATION. — 1.  Since  a  number  can  be  subtracted  only 
from  another  of  the  same  kind,  (Art.  43),  figures  of  the  same  name 
are  placed  in  column,  to  be  convenient  to  each  other,  the  less  number 
being  placed  below  as  a  matter  of  custom. 

2.  Commence  at  the  right  to  subtract,  so  that  if  any  figure  is 
greater  than  the  one  above  it,  the  upper  may  be  increased  by  10,  and 
the  next  lower  figure  by  1 ;  both  numbers  being  equally  increased, 
since  10  in  any  place  is  equal  to  1  in  the  next  higher  place. 

The  difference  between  two  numbers  is  the  same  as 
the  difference  between  those  numbers  increased  equally. 
Thus,  the  difference  between  2  and  5,  is  the  same  as  the 
difference  between  2  +  10  and  5  +  10,  or  12  and  15. 

There  is  another  method  of  performing   the  operatina 

RKTIKW. — 43.  What  is  subtraction  the  reverse  of?  Why?  What  ii 
necessary  in  order  that  one  number  may  be  subtracted  from  another  7 
Why?  44.  What  is  the  rule  for  subtracting  simple  numbers?  Explain 
the  example,  and  give  the  reasons  for  the  rule. 


21  RAY'S   HIGHER   ARITHMETIC. 


when  the  lower  figure  is  greater  than  the  upper,  called 
borrowing.  To  explain  this,  take  26  from  75. 

SOLUTION. — Since  6  units  can  not  be  taken  from  '  " 

5  units,  borrow  1  ten  from  the  7  tens,  and  add  it  to 
the  6,  which  will  make  16  units.     Then,  0  units  from          4  9 
15  units  leaves  9  units;  and  2  tens  from  6  tens,  (the 
number  left  in  tens'  place)  leaves  4  tens :  and  the  whole  remainder 
is  4  tens  and  9  units,  or  49. 

PROOF. — Add  the  remainder  to  the  less  number.     The 
sum  will  be  equal  to  the  greater. 
NOTE. — For  proof  by  casting  out  the  9's,  see  page  69. 

EXAMPLES    FOR  PRACTICE. 

1.  From  30020037  take  50009. 

When  there   are   no   figures   in   the  37 

lower  number  to  correspond  with  those  50009 

in  the  upper,  consider  the  vacant  places  29970028   Rem 
occupied  by  zeroes. 

FROM  TAKE  REM. 

2.  79685    .    .  30253    .    .    49432. 

3.  1145906    .    .  39876    .     1106030. 

4.  7150490    .    .    .       92367    .     7058123. 

5.  2900000    .    .         777888  '.     2122112. 

6.  71086540  .  .  64179730  .     6906810. 

7.  10143072  .  .  9247568  .       895504. 

8.  20286144  .  .  18495136  .     1791008. 

9.  25047361  .  .  7140851       17906510 

10.  101067800     .    .  100259063    .    .  808737. 
In  each  of  the  following  examples  to  the  18th,  subtract 

tne  first  number  from  the  second,  then  from  the  re- 
mainder, and  so  on,  till  a  remainder  is  found  less  than 
the  first  number.  LAST  REM 

11.  7359  and  36800 '.  .  "5! 

12.  7598  and  35768 5376. 

13.  59649  and  248697 10101. 

14.  730928  and  2500000 307216. 

15.  3956847  and  21018802  ....  1234567. 

16.  9080907  and  60000000  .  .  .  .  5514558. 

17.  1234567  and  10000000 123464. 

18.  938979  and  9389790  0. 


SUBTRACTION. 


19.  A  merchant  has  $3050;   what  sum  will  he  have, 
after  paying  $2364?  An*.  $686. 

20.  How  many  years  from  the   discovery  of  America 
in  1492,  to  its  Independence  in  1776?  Ans.  284. 

21.  A  has  $5000,  and  owes  $2705;  what  will  be  left 
after  paying  his  debts?  Ans.  $2295. 

22.  A   farm    that  cost   $7253   was   sold    at   a   loss   of 
$395;  for  how  much  was  it  sold?  Ans.  $6858. 

23.  What   number   must  be   added   to   6428    that   the 
«um  shall  be  8350?  Ans.  1922. 

24.  The  difference  of  two  numbers  is  19034,  and  the 
greater  is  75421;  what  is  the  less?  Ans.  56387. 

25.  How    many    times    can    285    be    subtracted    from 
1425?  Ans.  5. 

26.  Which  is  the  nearer  number  to  920736;  1816045 
or  25427?  Ans.  Neither.     Why? 

27.  Of  the  numbers  23452  and  60000,  how  much  is 
one  nearer  to  41386  than  the  other?  -4ns.  680. 

28.  A  merchant  owing  $5000,  paid  $2175,  and  $1850; 
how  much  was  unpaid?  Ans.  $975. 

29.  From  a  tract  of  land  containing  10000  acres,  the 
owner  sold   to   A   4750   acres;  and  to  B  875  acres  less 
than  A:   how  many  acres  had  he  left.  Ans.  1375. 

30.  A,  B,  C,  D,  are  4  places  in  order  in  a  straight  line. 
From  A  to  D  is  1463  miles;  from  A  to  C,  728  miles; 
and  from  B  to  D,  1317  miles.     How  far  is  it  from  A  to 
B,  from  B  to  C,  and  from  C  to  D? 

Ans.  A  to  B,  146  miles;  B  to  C,  582;  C  to  D,  735. 

31.  From  Pittsburgh   to   New  Orleans  is  1999  miles; 
from   Pittsburgh    to   Memphis   is  1260  miles ;    and   from 
Cincinnati   to  New   Orleans  is  1523  miles:  how  far  is  it 
from  Cincinnati  to  Pittsburgh,  from  Cincinnati  to  Mem- 
phis, and  from  Memphis  to  New  Orleans? 

Ans.   From   C.  to   P.  476   miles;   from  C.  to  M.  784- 
from  M,  to  N.  0.  739  miles. 

ART.  45.    EXAMPLES  IN  ADDITION  AND  SUBTRACTION. 

1.  Add  37452  to  19576,  and  from  the  result  subtract 
the  sum  of  19914  and  19715.  Am.  17399. 

RE  v  i  B  w.  — 44.  What  is  the  method  of  borrowing?    What  is  the  proof 

for  subtraction  't 

3 


¥8  RAY'S   HIGHER   ARITHMETIC. 

2.  From  the  sum  of  28374  and   56132,  subtract   the 
differed  of  I'lOOOO  and  84506.  Am.  69012. 

3.  To  the  difference  of  95603  and  44571,  add  the  dif- 
ferent of  8632  and  4179.  AM.  55485. 

4.  From  the  difference  of  7325  and  3149  subtract  the 
difference  of  8645  and  4702.  Ans.  233. 

5.  From  what  number  must  256  be  subtracted  to  leave 
the  remainder  144?  Am.  400. 

6.  To  what  number  must  1327  be  added,  to  make  the 
sum  5000?  Am.  3673. 

7.  From  what  number  must  27  be  subtracted  3  times, 
to  leave  38?  Am.  119. 

8.  To  what  number  must  65  be  added  4  times,  to  make 
the  sum  297?  Am.  37. 

9.  The  distance  by  railroad  from  Cincinnati  to  Cleveland 
is  275  miles.    When  one  locomotive  is  60  miles  from  Cin- 
cinnati, and  another  118  miles  from  Cleveland,  how  far  are 
they  apart?      How  far  apart,  when  one  has  traveled  174 
miles,  and  the  other  183  miles? 

Ans.  97  miles  and  82  miles. 

10.  A  was  37  years  old  when  B  was  born;  how  old  was 
A,  when  B  was  25  years  of  age?  and  how  old  was  B,  when 
A  was  73  years  of  age?  Ans.  62  and  36. 

11.  Which  is  the  nearer  number  to  356908;  713866 
or  7849,  and  how  much?  Ans.  The  last,  by  7899. 


V.   MULTIPLICATION. 

ART.  48.  MULTIPLICATION  is  taking  one  number  as 
~  times  as  there  are  units  in  another;  or 

Multiplication  is  a  short  method  of  adding  numbers  that 
•re  equal. 

The  number  to  be  taken,  is  called  the  Multiplicand;  th« 
uther  number,  the  Multiplier;  and  the  result  obtained,  the 
Product. 


RMVIKW. — 46.  What  is  Multiplication  ?     What  is  the  Multiplicand? 


MULTIPLICATION. 


27 


How  many  trees  in  3  rows,  each  containing  42  trees. 

SOLUTION. — Since  3  rows  contain  3     First  row,      42  trees. 
times  as  many  trees  as  one  row,  take  42 
three  times.    This  may  be  done  by  writing 
42  three  times,  and  then  adding.     This 
gives  126  for  the  whole  number  of  trees. 

Instead,  however,  of  writing  42  three  times,  write 
t  once ;  then  placing  under  it  the  figure  3,  the  number 


Second  row    42  trees. 
Third  row      42  trees. 

126  tree*. 


trees. 


„  times  it  is  to  be  taken,  say,  3  times  2  are  6,  and  3     - 

times  4  are  12.     This  process  is  Multiplication.  l^O  trees. 

The    Multiplicand    and    Multiplier   are    together  called 
Factors  (makers),  because  they  make  the  product. 

MULTIPLICATION  TABLE. 


o 

5 

r. 

-) 

cr. 

en 

*>. 

CO 

to 

1  —  1 

a 

eg 

gc 

-j 

—  . 

en 

4- 

cc 

to 

P_I 

1 

CO 
X 

CO 

CO 

4- 

co 

10 

co 

0 

10 
X 

to 

0 

to 

4- 

to 

to 

to 

p—  • 

00 

a 

~ 

— 

to 

H 

X 

Cs 

to 

en 

0 

en 

en 
£ 

C'' 

1r 

i^ 

4- 

to 

co 

0 

00 

cr. 

Co 
CO 

CO 

to 

to 

4- 

to 

p—  •• 
gc 

Oi 

to 

0 

Cr. 

CO 

50 

S 

~j 

-i 
to 

C-. 

g 

§ 

en 

a 

CT 

to 

4- 
X 

f- 

4- 

1 

Co 

CO 

to 

to 

1^0 

,!i 

0 

r— 

to 

• 

1  —  1 

8 

-~ 

5' 

-/- 

C.-' 

V 

2 

-i 

2 

a 

& 

Ci 

O 

Oi 

en 

O 

i 

£ 

C- 

co 

C 

to 

K 

M 
Gjn 

C 

01 

;   ^" 

—  ' 
—  ' 

4- 

i  — 
| 

0 

to 

~ 

-r 

90 

4- 

-i 

00 

-I 

to 

0 

c; 

g 

K 

f3 

CO 

c. 

g 

to 

4- 

CO 

•-• 

10 

o> 

i 

co 

CO 

i  •- 

r. 

rr 

~ 

p 

0 
X 

2 

00 
4- 

:! 

c- 

0 
CO 

C£> 

t; 

1 

CO 

g 

t: 

^ 

^, 

a 
5 

c.-i 
to 

t 

CO 
0 

1  V 

-j 

to 

S 

rt 

£ 

(fi 

0 

QC 

TC 

X 

r. 

to 

£ 

en 

c- 

v: 

i 

:o 

1C 

| 

'H 

GO 

-f 
5 

-  1 

1  —  ' 

OS 

to 

en 

CO 

>u 

CO 

C,i 

l_0 

h-  ' 

- 

X 

0 

0 
0 

OC 

i  —  ' 

t  0 

w 

Z- 

1 

Cti 

a 

t  c 

y 

« 

to 

5 

X 

o 

-_I 

c-. 

c,-. 

p 

t 

EC 

O 

1 

§ 

i; 

0 

i 

-I 

~ 

i 

en 

or 

i 

CO 

C 

t  -. 

•- 

•^ 
O 

!| 

to 

1 

-0 

X 

-: 

-I 

0 

a 

en 

4- 

i 

CO 

to 

to 

p^ 

vr 

90 

S 

-  1 

a 

a 

en 

t 

CO 
0- 

to 
to 

^ 

to 

1 

10 

i 

to 

1  0 

0 

4- 

t— 

--2 
10 

,  —  i 
ao 

C' 

i—  ' 

a 

X 

!  

0- 

Ci 

1— 

t 

1  —  • 
CO 
10 

± 

~- 

5 

0 

oo 

4- 

to 

c-. 
C 

^ 

CO 

to 

4- 

t—  ' 
to 

i 

0 

to 

1 

to 

IO 

to 

'_? 

00 

<  — 
CM 

X 

to 

i 

CM 

a 

4- 
0- 

—  i 

i  — 
-  — 

1  — 

-P 

^ 

X 

f; 

o 

to 

CO 

10 

c. 

1—  I 
CO 

to 
/ 

- 

to 

•£- 

Li 

10 

10 

0: 

y 

to 

1C 

to 

•—  ' 

§ 

X 

to 

y 

O-i 

4- 

t 

\-^ 

cr- 

'  — 
to 

-_~ 

V. 

y 

4- 

— 

c-! 

i1: 

i 

>—  • 

•b. 

1 

to 
/. 

0-1 

0 

to 
in 

Oi 

to 

4- 

O 

to 
to 

Oi 

to 

r— 

0 

1 

3C 

C- 
& 

CJt 

CO 

en 

1  •- 
C 

~ 

CQ 

O 

ft 

a 

c 

| 

CO 

O 

^ 

i 

CO 

1 

z 

4- 

co 

cr 

00 
X 

M 

Q 

-  1 

10 

to 

/. 

0 

'cr. 

—  . 

3 

to 

!u 

0 

1 

Ci 

to 

4- 

t  '-• 

CO 

r. 

1C 

to 

—  ' 

5 

to 

1 

p—  ' 
0( 

a 

r 
-  1 

5 

in 

r' 

!  '_ 

y 

a 

rr 

cr- 

0 

t: 

at 

- 

x 

a 

X 

vc 

en 

i  —  i 

:o 
to 

H 

4- 

Qa 

H* 

co 
8 

CO 

fe 

•co 

CO 

S 

~- 

to 

-/ 

S- 

1 

t"o 

ii 

to 

1— 
cr. 

i  —  ' 

tc 

V. 

^ 

a 

10 

t 

i  — 

r. 

"r 
OJ 

S 

-  1 
t: 

en 

4- 

CO 

1—  i 

oo 

CO 

-f 

CO 

0. 
1  —  ' 

co 
i"c 

CO 

to 

1 

V. 

10 

i 

-"i 

i 

X 

i 

2 

| 

-  1 

to 

S 

-- 

§ 

-i 

a 

C'l 

U 

S 

i 

CO 
00 
O 

co 
a 

CO 

B 

r 

1 

~ 

l_0 

5 

i 

1 

t  -j 
S 

-/ 
o 

I 

—  ' 

i 

I 

S 

8 

£ 

g 

28  RAY'S   HIGHER   ARITHMETIC. 


ART.  47.  THEOREM. —  The  product  of  two  numbers  is  tht 
tame,  whichever  factor  is  the  multiplier. 

DEMONSTRATION. — To  find  the  number  of  stars  in  this  diagram, 
multiply  the  number  of  stars  in  a  row,  by  the  number  of  rows.     Count- 
ing up  and  down,  there  are  4  in  a  row,  and  3  rows;  . 
hence,  there  must  be  3  times  4,  =  12  stars.     But 
across,  there  are  3  in  a  row,  and  4  rows;   hence, 
there  must  be  4  times  3,  =  12  stars.     The  two  pro- 
ducts, 3  times  4  and  4  times  3,  are  equal,  since 
each  gives  the  whole  number  of  stars  in  the  diagram.     The  same 
may  be  shown  of  any  two  numbers. 

ART.  48.  THEOREM. — The  product  is  always  of  the  same 
name  as  the  multiplicand. 

DEMONSTRATION. — The  nature  of  a  number  is  not  changed  by 
repeating  it.  Thus,  5  dollars  multiplied  by  2,  that  is,  taken  twice, 
must  be  10  dollars. 

ART.  49.  THEOREM. — The  multiplier  must  always  be  an 
abstract  number. 

DEMONSTRATION. — The  multiplier  shows  the  number  of  times 
the  multiplicand  is  to  be  taken ;  hence,  it  can  not  be  yards,  bushels, 
or  any  concrete  number.  We  can  attach  no  idea  to  taking  any  thing 
PO  many  dollars  times,  or  so  many  inches  times.  It  may  then  be  asked, 
how  explain  the  solution  of  such  questions  as  "  What  will  3  yards 
of  cloth  cost  at  5  dollars  a  yard  ?"  Ans.  By  the  following  or  s  simi- 
lar Analysis :  Three  yards  will  cost  three  times  as  much  as  one  yard ; 
therefore,  if  one  yard  cost  5  dollars,  3  yards  will  cost  3  times  5  dol- 
lars, =  $15. 

REMARK. — Hence,  it  is  absurd  to  talk  of  multiplying  dollars  by 
dollars.  It  would  be  as  rational  to  propose  to  find  the  product  of 
8  apples  by  2  turnips. 

ART.  50.    Multiplication  is  divided  into  two  cases: 

1.  When  the  multiplier  does  not  exceed  12. 

2.  When  the  multiplier  exceeds  12. 

RULE  WHEN  THE  MULTIPLIER  DOES  NOT  EXCEED  12. 

1.  Write  the  multiplicand;  place  the  multiplier  under  it,  and 
diaw  a  line  beneath. 

2.  Begin  with  units;  multiply  each  figure  of  the  multiplicand 
by  the  multiplier,  setting  down  and  carrying  as  in  Addition. 


MULTIPLICATION. 


At  the  rate  of  53  miles  an  hour,  how  far  will  a  railroad 
car  run  in  four  hours? 

SOLUTION. — Here   flay,  4  times  3  (units)  are          53  miles. 
12  (units) ;  write  the  2  in  units'  place,  and  carry  4 

the  1  (ten) ;  then,  4  times  5  are  20,  and  1  carried       o^To 
makes  21  (tens),  and  the  work  is  complete. 

DEMONSTRATION. — The  multiplier  being  written  under  the  mul- 
tiplicand for  convenience,  begin  with  units,  so  that  if  the  product 
should  contain  tens,  they  may  be  carried  to  the  tens ;  and  so  on  for 
each  successive  order. 

Since  every  figure  of  the  multiplicand  is  multiplied,  therefore,  the 
whole  multiplicand  is  multiplied. 

PROOF. — Separate  the  multiplier  into  any  two  parte; 
multiply  by  these  separately.  The  Bum  of  the  products 
must  be  equal  to  the  first  product. 

EXAMPLES  FOR    PRACTICE. 

MULTIPLICAND.  MULTIPLIER.  PRODUCT. 

1.  195 3 585. 

2.  3823   ....  4 15292. 

3.  8765   ....  5 43825. 

t.  98374  ....  6 590244. 

o.  64382  ....  7 450674. 

6.  58765  ....  8 470120. 

7.  837941.  ...  9  ....  7541469. 

8.  645703 ....  10  ....  6457030. 

9.  407649 ....  11  ....  4484139. 

10.  If  4  men  can  perform  a  certain    piece  of  work  in 
15  days,  how  long  will  it  require  1  man? 

SOLUTION. — One  man  will  be  four  times  as  long  as  four  men. 
4  X  15  =  60  days. 

11.  How  many  pages  in  a  half-dozen  books,  each  con- 
taining 336  pages?  Ans.  2016. 

12.  How  far  can  an  ocean  steamer  travel  in  a  week,  at 
he  rate  of  245  miles  a  day?  Ans.  1715  miles. 


REVIEW. — 46.  Repeat  tho  Multiplication  Table.  47.  Prove  the  theo- 
rem. 48.  How  is  the  denomination  of  the  product  known  ?  Why  t 
49.  What  kind  of  a  number  is  the  multiplier?  Why?  50.  How  is  multi- 
plication divided  ?  What  is  the  rule  when  tho  multiplier  Joes  not  exceed 
12?  Give  the  reason.  What  is  the  proof  ? 


30  RAY'S   HIGHER   ARITHMETIC. 


13.  What   is   the   yearly  expense   of  a   cotton-mill,   if 
$32053  are  paid  out  every  month?         Ana,  $384636. 

ART.  51.   RULE  WHEN  THE  MULTIPLIER  EXCEEDS  12. 

1.  Write  the  multiplier  under  the  multiplicand,  placing  figuret 
of  the,  same  order  under  each  other. 

2.  Multiply  by  each  figure  of  the  multiplier  successively ;  fill 
by  the  units1  figure,  then  by  the  tens'  figure,  &c.;  placing  the  righi 
hand  figure  of  each  product  under  that  figure  of  the  multiplier 
which  produces  it. 

3.  Add  the  several  partial  products  together;  their  sum  will  be 
the  required  product. 

Multiply  246  by  235. 

SOLUTTON.  —  First  multiply  by  246 

6  (units),  and  place  the  first  figure  235 

of  the  product,  1230,  under  the  5  -i  9  o  A 

(units).    Then  multiply  by  3  (tens),  i  {j  £  U  product  by         0 

arid   place   the  first  figure  of   the  product  by     d  0 

product,  738,  under  the  3  (tens).  *  JZ  product  by  IQV 
Lastly,  multiply  by  2  (hundreds),  57810  product  by  2  3  5 
and  place  the  first  figure  of  the 

product,  492,  under  the  2  (hundreds).     Then  add  these  several  pro- 
ducts for  the  entire  product. 

ANALTHIS. — The  0  of  the  first  product,  1230,  is  units,  Art.  60. 
The  8  of  the  2d  product,  738,  is  tens,  because  3  (tens)  times  6  =  6 
times  3  (tens)  =18  (tens);  giving  8  (tens)  to  be  written  in  the 
tens'  column.  The  2  of  the  3d  product,  492,  is  hundreds,  because  2 
(hundreds)  times  6=6  times  2  (hundreds)  =  12  (hundreds),  giving 
2  (hundreds)  to  be  written  in  the  hundreds'  column.  The  right 
hand  figure  of  each  product  being  in  its  proper  column,  the  other 
figures  will  fall  in  their  proper  columns ;  and  each  line  being  the 
product  of  the  multiplicand  by  a  part  of  the  multiplier,  their  sum 
will  be  the  product  by  all  the  partt  or  the  whole  of  the  multiplier. 

METHODS  OP   PROOF. 

1.  Multiply   the   multiplier  by  the   multiplicand;    thii 
product  must  be  the  same  as  the  first  product. 

2.  The  same  as  when  the  multiplier  does  not  exceed  12. 
REMARK  — For  proof  by  casting  out  the  9's,  see  Art.  100. 

R  E  v  i  K  w . — What  i«  the  rule  when  the  multiplier  exceeds  12  ? 


CONTRACTIONS  IN   MULTIPLICATION.  3] 


ART.   52.    Although    it    is  £46 

customary  to  use   the   figures 

of  the   multiplier  in   regular  r- 

order  beginning  with  units,  it  738     product  by     30 

will  give  the  same  product  to  ^  ?        product  by  200 

use  them  in  any  order,  observ-  1230  product  by 5 


ig  that  the  right-hand  figure      57810  product  by  2  3  5 
f  each  partial  product  must  be 
faced  under  the  figure  of  the  multiplier  which  produces  it. 

1. 
2. 
3. 

EXAMPLES  FOR 

7198x216 

PRACTICE. 

.    .    .    .  —1554768. 

8862  x  189 

.    .    .    .  —1674918. 

7575x7575     . 

.    .    .     =57380625. 

4. 

15607  X  3094     . 

•4-  i;  .     =48288058. 

5. 

93186  x  4455     . 

.    .    .  =  415143630. 

6. 

135790x24680   . 

.    .      =3351297200. 

7. 

3523725x2583      . 

.    .     =9101781675. 

8. 

4687319x1987     . 

.    .      =9313702853. 

9. 

9264397X9584     . 

.    .    =88789980848. 

10. 

9507340x7071     . 

.     .    =67226401140. 

11. 

1644405  X  7749     . 

.    .   =12742494345. 

12. 

1389294x8900     . 

.    .   =12364716600. 

13. 

2778588x9867     . 

.  '  ;   =  27416327796. 

14. 

204265x562402. 

.      =  114879044530. 

15. 

39948123x6007     . 

.      =  239968374861. 

16. 

57902468  X  5008      . 

.      =  289975559744. 

17. 

57902468  X  5080      . 

.      =  294144537440. 

18. 

3081025X2008036 

.    =6186809116900. 

19. 

860389657x96795   . 

.  =  83281416849315. 

CONTRACTIONS  IN  MULTIPLICATION. 

CASE  I. 
WHEN  THE  MULTIPLIER  18  A  COMPOSITE   NUMBER. 

ART.  53.  A  composite  number  is  the  product  of  tiro  or 
more  whole  numbers,  each  greater  than  1,  called  its  fac- 
tors. Thus,  10  is  a  composite  number,  whoso  factors  arc 
2  and  5:  and  30  is  one  whose  factors  are  £.  3  and  5- 


32  RAY'S   HIGHER   ARITHMETIC. 


RULE. — Separate  the  multiplier  into  two  or  more  factors.  Mul 
tiply  first  by  one  of  the  factors,  then  this  product  by  another 
factor,  and  so  on  till  each  factor  has  been  used  as  a  multiplier. 
The  last  product  will  be  the  entire  product  required. 

At  7  cents  a  piece,  what  will  6  melons  cost? 

ANALYSIS. — Three  times  2  OPERATION. 

times  are  6  times.     Hence,  it  7  cents,  cost  of  1  melon. 

Is  (he  same  to  take  2  times  7,  2 

and  then  take  this  product  3  ~TT~ 

times,  as   to  take   6  times  7.  l 1  Cents'  C08t  of  2  melon8' 

And  the  same  may  be  shown  of          

any  other  composite  number.  42  cents,  cost  of  6  melons. 

EXAMPLES    FOR    PRACTICE. 

1.  At  the  rate  of  37  miles  a  day,  how  far  will  a  man 
walk  in  28  days?  Ans.  1036  miles. 

2.  Sound  moves  about  1130  feet  per  second  :  how  far 
will  it  move  in  54  seconds?  Ans.  61020  feet. 

3.  Multiply  9765  by  35.  Ans.  341775. 

CASE  II. 

WHEN    THE  MULTIPLIER   IS   ONE    WITH    CIPHERS    ANNEXED, 

AS  10,  100,  1000,  &c. 

ART.  54.  RULE. — Annex  to  the  multiplicand  as  many  ciphers 
as  there  are  in  the  multiplier;  the  result  will  be  the  required 
product. 

DEMONSTRATION. — By  the  principles  of  Notation,  (Art.  26), 
placing  one  cipher  on  the  right  of  a  number,  changes  the  units  into 
tens,  the  tens  into  hundreds,  and  so  on,  and  therefore,  multiplies  the 
number  by  10. 

Annexing  two  ciphers  to  a  number  changes  the  units  into  hun- 
dreds, the  tens  into  thousands,  and  so  on,  and,  therefore,  multiplies 
the  number  by  100.  Annexing  three  ciphers  multiplies  the  number 
by  1000,  &c. 

EXAMPLES   FOR  PRACTICE. 

1.  Multiply  375  by  100 Ans.     37500. 

2.  Multiply  207  by  1000 Ans.  207000. 


REVIEW. — 52.  Explain  the  Example.  53.  What  is  a  composite  number? 
What  are  its  factors?  What  is  the  rule  for  multiplying  by  a  ccxnposiie 
camber  ? 


OONTKACTIONS  IN   MULTIPLICATION.  33 


CASE  III. 

WHEN  CIPHERS  ARE  AT  THE  RIGHT  OF  ONE  OR  BOTH 
FACTORS. 

ART.  55.  RULE. — Multiply  as  if  there  were  no  ciphers  at  *A< 
fight  of  the  numbers;  then  annex  to  the  product  as  many  ciphers 
«.«  there  are  at  the  right  of  both  the  factors. 

Find  the  product  of  5400  by  130.  5400 

130 

SOLUTION. — Find  the  product  of  54  by  13,        162 
and  then  annex  three  ciphers,  the  number  at  the        5  4 
right  of  both  factors. 


702000 

ANALYSIS. — Since  13  times  64  =  702,  it  follows  that  13  times  64 
hundreds  (6400)  =  702  hundreds  (70200);  and  130  times  6400=10 
times  13  times  5400  =  10  times  70200  =  702000. 

OR  TITOS:    The  operation  is  the  same  as  that 

in  the  Rule  for  Multiplication,  (Art.  51),  except  

that  the  ciphers  on  the  right  are  omitted  until  162000 

the  sum  of  the  partial  products  is  found.     This  is  5400 

evident  from  the  operation  in  the  margin.  >7fl9ft  ftn 

EXAMPLES  FOR  PRACTICE. 

1.  15460  X  3200          .    .    .    .  =  49472000. 

2.  30700x5904000    .    .  =181252800000. 

3.  67051800x37800.     .      =2534558040000. 

4.  9730000x645100      .      =6276823000000. 

5.  46218000x757000      .    =34987026000000. 

6.  9800100x6514000    .    =63837851400000. 

CASE   IV. 

WHEN    THE    MULTIPLIER    IS    A    LITTLE    LESS    THAN 

10,  100,  1000,  &0. 

ART.  56.  RULE. — Annex  to  the  multiplicand  as  many  ciphers 
as  the^e  are  figures  in  the  multiplier,  and  from  the  result  subtract 
the  pr  iduct  of  the  multiplicand  by  the  number  the  multiplier  icania 
of  being  10,  100,  1000,  &c. 

REVIEW. — 54.  What  is  the  rule  for  multiplying  by  1  with  cipher! 
annexed?  Give  the  reasons.  55.  What  is  the  rule  for  multiplying  when 
ciphers  are  at  the  right  of  one  or  both  factors?  Explain  the  example. 


34  RAY'S   HIGHER   ARITHMETIC. 


Multiply  3046  by  997.  OPERATE. 

ANALYSIS.— Since  997  is  the  same   as  3046 

1000  diminished  by  3,  to  multiply  by  it  is  997 

the  same  as  to  multiply  by  1000,  (that  is,  ^046000 

to  annex  3  ciphers),  and  by  3,  and  take  the  Q  -.  q  o 

difference  of  the  products,  which  agrees  with 
the  rule;  and  the  same  can  be  shown  in  any 
•imilar  case. 

EXAMPLES  FOR  PRACTICE. 

1.  7023x99 =  695277. 

2.  16642x996 =16575432. 

3.  372051  X  9998 =  3719765898. 

NOTE. — If  there  are  ciphers  at  the  r;ght  of  one  or  both  the  'ac- 
tr.  re,  the  rule  may  still  be  applied,  (/.it.  65.) 

4.  642438x93 =59746734. 

5.  28714X995 =28570430. 

6.  99999x99991     .    .    .    .=9999000009. 

7.  525734x9994.    .    .    .      =5254185596. 

8.  4127093x9989  ....    =41225531977. 

9.  2634527x9980  ....    =26292579460. 

10.  372918x9999600     .     =3729030832800. 

11.  203800x9997000     .    =2037388600000. 

12.  811876x99988     .    .    .    =81177857488. 

13.  3606253x9999990     .=36062493937470. 

14.  2055416X992     ....      =2038972672. 

CASE  v. 

ART.  57.  Another  mode  of  contraction,  of  frequent  \jse, 
is  comprised  in  the  following 

RULE. — Derive  the  partial  products,  when  possible,  from  each 
other,  commencing  with  that  figure  of  the  multiplier  which  is  most 
convenient  for  the  purpose;  making  use  of  two  or  more  figures  of 
the  multiplier  at  once,  when  it  can  be  done,  and  setting  the  right- 
hand  figure  of  each  partial  product  under  the  right-hand  figure 
of  the  multiplier  in  use  of  the  time. 


REVIEW. — 56.  What  is  the  rule  for  multiplying  by  a  number  that  wants 
but  little  of  100,  1000,  Ac.?  Explain  the  example.  If  there  are  ciphers  at 
the  right  of  one  or  both  factors,  what  may  be  done  ?  57.  What  other 
method  is  there  of  contracting  multiplication  ? 


DIVISION   OP   SIMPLE  NUMBERS.  35 

Multiply  387295  by  216324. 

SOLUTION. — Commence  with  the  3  of  QQfronti 

tie  multiplier,  and  obtain  the  first  partial  9 1  I SOA 

product,  1161885;  then  multiply  this  pro-  wlbd24 

duct  by  8,  which  gives  the  product  of  the  1  1  G  1  8  8  5 

multiplicand  by  24  at  once,  (since  8  times  9295080 

8  times  any  number  make  24  times  it.)      83655720 
Set  the  right-hand  figure  under  the  right-      ™7ft1  9  no  ron 
hanJ   figure   4  of   the  multiplier  in  use. 

Multiply  the  second  partial  product  by  9,  which  gives  the  product 
of  the  multiplicand  by  216,  (since  9  times  24  times  a  number  make 
216  times  that  number.)  Set  the  right-hand  figure  of  this  partial 
product  under  the  6  of  the  multiplicand;  and,  finally,  add  to  obtain 
the  total  product. 

EXAMPLES  FOR  PRACTICE. 

1.  38057x48618.    .'    .    .    .    =1850255226. 

2.  '267388x14982 =4006007016. 

3.  481063x63721  ....      =30653815423. 

4.  66917X849612     .    .    .      =56853486204. 

5.  102735x273162  .  .  .  =28063298070. 

6.  536712x729981  .  .  .  =391789562472. 

7.  750764x315135  .  .  .  =236592013140. 

8.  930079x251255  =233686999145. 


VI.   DIVISION. 

ART.  58.  DIVISION  is  the  process  of  finding  how  many 
times  one  number  is  contained  in  another. 

Also,  Division  is  the  process  of  finding  one  of  the 
factors  of  a  given  product,  when  the  other  is  known. 

The  number  contained  in  the  other  is  the  Divisor,  the 
other  number  is  the  Dividend,  and  the  result  obtained  is 
the  Quotient. 

The  Remainder  is  the  number  which  is  sometimes  left 
after  dividing. 

Thus,  in  the  question,  How  often  is  2  contained  in  7?  the  divisor 
is  2,  the  dividend  7,  the  quotient  3,  and  the  remainder  1. 

NOTE. — Dividend  signifies  to  be  divided.  Quotient  is  derived  from 
the  Latin  word  quoties,  which  signifies  how  often. 


36  RAY'S   HIGHER   ARITHMETIC. 

ART.  59.  The  divisor  and  quotient  in  Division,  cor 
respond  to  the  factors  in  Multiplication,  and  the  dividend 
corresponds  to  the  product.  Thus: 

FACTORS.     PRODUCT. 

MULTIPLICATION.     .     3X5  =  15. 

DIVIDEND.  DIVISOR.    QUOTIENT. 

DIVISION.       ...      1 5  divided  l>y   3     =     5 
or,   1 5  divided  by  5     =     3. 
There  are  three  methods  of  expressing  division  ;  thus, 

12-j- 3,  **        or       3)12. 

Each  indicates  that  12  is  to  be  divided  by  3. 

ART.  60.  THEOREM. — Division  is  a  short  method  of 
making  several  subtractions  of  the  same  number. 

DKMONSTRATION. — To  prove  this,  find  how  many  times  2  cents 
is  contained  in  7  cents. 

Two  cents  from  7  cents  leaves  5          7  cents, 
cents ;  2  cents  from  5  cents  leaves          2  cents  contained  once. 

3   cents ;    2   cents   from   3   cents 

t/  cents* 
leaves  1  cent.  O  *  •     j  O  A- 

4  cents  contained  4  times. 
Here,  2  cents  is  taken  3  times         — 

from  (out  of )  7  cents,  and  1  cent          3  cents. 

remains  ;   hence,  2  cents   is   con-          "  cents  contained  o  times. 

tained  in  1  cents  3  times,  with  a          \  cent  \efa 

remainder  of  1  cent. 

COROLLARIES. 

COR.  1.  The  Dividend  and  Divisor  must  always  be  of 
the  same  denomination. 

COR.  2.  The  quotient  is  always  an  abstract  number; 
merely  showing  how  many  times  the  divisor  is  contained  in 
the  dividend. 

COR.  3.  The  remainder  is  always  of  (he  same  name  ai 
the,  dividend. 

Multiplication  is  a  short  method  of  making  several  iddi- 
tions  of  the  same  number ;  Division  is  a  short  method  of 
jaaking  several  subtractions  of  the  same  number :  hentv?, 
Division  is  the  reverse  of  Multiplication. 


REVIEW.  — 67.  Explain  the  example.  58.  What  is  Division  ?  What 
\s  the  Divisor?  Dividend?  Quotient?  Remainder?  What  Joes  dividend 
mean?  What  does  quotient  mean?  59.  If  the  dividend  is  the  product, 
what  are  the  factors?  If  the  divisor  and  quotient  are  factors,  what  ii 
the  product?  What  are  the  signs  of  Division? 


DIVISION   OP   SIMPLE   NUMBERS.  37 


The  next  stop  is  to  learn  tho  Division  Table,  which 
gives  the  quotient  in  all  cases  where  the  divisor  and  quo- 
tient are  both  12  or  less,  and  there  is  no  remainder.  It 
is  made  from  the  Multiplication  Table.  Thus,  4  in  12  is 
contained  8  times,  because  3  times  4  are  12. 

ART.  61.  When  the  divisor  does  not  exceed  12,  the 
operation  is  called  Short,  Division. 

RULE  FOR  SHORT  DIVISION. 

1.  Write  the  dioisor  on  the  left  of  the  dividend  with,  a  curved 
line  between  them.     Begin  at   the  left,  divide  successively  each 
figure  of  the  dividend  by  the  divisor,  and  set  the  quotient  beneath 
the  figure  divided. 

2.  Whenever  a  remainder  occurs,  prefix  it  to  the  figure  in  the 
next  lower  order,  and  divide  as  before. 

3.  If  the  figure  in  any  order  does  not  contain  the  divisor,  place 
a  cipher  beneath  it,  prefix  it  to  the  figure  in  the  next  lower  order, 
and  divide  as  before. 

4.  If  there  is  a  remainder  after  dividing  the  last  figure,  plact 
the  divisor  under  it  and  annex  it  to  the  quotient. 

How  often  ia  2  cents  contained  in  652  cents? 

SOLUTION  . — Two  in  6  (hundreds)  is  contained  3        2)652 
(hundreds)  times;  2  in  5  (tens)  is  contained  2  (tens)  Q~o7? 

times,  with  a  remainder  of  1  (ten);  lastly,  1  (ten) 
prefixed  to  2  makes  12,  and  2  in  12  (units)  is  contained  6  times, 
making  the  entire  quotient  32G. 

DEMONSTRATION. — Commence  at  the  left  to  divide,  so  that  if 
there  is  a  remainder  it  may  be  carried  to  the  next  lower  order. 

By  the  operation  of  the  rule,  the  dividend  is  separated  into  parts 
corresponding  to  the  different  orders.  Having  found  the  number  of 
times  the  divisor  is  contained  in  each  of  these  parts,  the  sum  of  these 
must  give  the  number  of  times  the  divisor  is  contained  in  the  wholt 
dividend.  Analyze  the  preceding  dividend  thus : 

652  =  600  +  40  +  12 

2  in  600  is  contained  300  times. 
2  in      4  0  is  contained      20 
2  in      1 2  is  contained          6 

Hence,  2  in  6  5  2  is  contained  326  times. 


RE  VIE  w. —  60.  Division  is  a  short  method  of  performing  what  opera- 
tion 1    Ezplain  the  example.     What  is  division  tho  reverse  of?    why  ? 


146022 


38  RAY'S   HIGHER   ARITHMETIC. 


METHODS  OF   PROOF. 

1.  Multiply  the  quotient  by  the  divisor,  and  add  in  the 
remainder,  if  any;  the  result  should  equal  the  dividend;  or, 

2.  Subtract  the  remainder,  if  any,  from  the  dividend ; 
the  result  divided  by  the  quotient,  should  give  the  first 
divisor. 

ART.  62.    It  has  been  shown,  (Art.  60),  that  the  divisor 

nd  dividend  must  be  of  the  same  denomination,  and  that 

thf  quotient  is  always  an  abstract  number.     It  is  now  to 

be  shown  how,  according  to  these  principles,  the  solution 

of  questions  in  Division  can  be  explained. 

All  questions  in  Division  belong  to  two  classes: 
I.  To  divide  a  number  in'-)  parts  each  containing  a  certain 

number  of  units,  and  to  find  the  number  of  parts. 

II.     To  divide  a   number  into  a  certain  number  of  equal 

parts,  and  to  fit  ul  (he  number  of  units  in  each. 

,  KXAMPLES  OF  THE    FIRST  CLASS. 

1.  At  $3  each,  how  many  hats  can  I  buy  for  $15? 
SOLUTION. — Since  1  hat  costs  $3.  I  can  buy  a  bat  as  often  as  $3 

is  contained  in  $15;  $3  in  $15  is  contained  5  times;  therefore,  I  can 
buy  6  hats. 

2.  At  $5  a  yard,  how  much  cloth  can  I  buy  for  $35? 

3.  In   a   school  of  60   pupils,  how  many  classes  of  10 
pupils  each? 

4.  I  can  walk  6  miles  in  1  hour:  in  how  many  hours 
can  I  walk  54  miles? 

EXAMPLES  OF  THE  SECOND  CLASS. 

5.  A  man  has  20  apples  to  divide  equally  among  4  boys: 
how  many  must  he  give  to  each? 

SOLUTION. — To  give  each  boy  1  apple  will  require  4  apples; 
hence,  each  boy  will  receive  one  apple  as  often  as  4  apples  are  con- 
fined in  20  apples;  4  apples  are  contained  in  20  apples  5  times; 
herefore,  each  boy  will  receive  5  apples. 

REVIEW.  — 60.  Of  what  denomination  must  the  divisor  bo  ?  the  remain- 
der? What  kind  of  number  is  the  quotient?  why?  61.  Wh;it  is  Short 
Division?  Give  the  rule.  Explain  the  example.  What  is  the  proof? 
2d  method  ?  How  is  the  remainder  written  ?  62.  What  two  classes  ol 
questions  arc  performed  by  division  ? 


DIVISION    OF   SIMPLE   NUMBERS.  39 

REMAKE. — The  solution  of  this  question  evidently  requires  that 
the  20  apples  should  be  divided  into  4  equal  parts,  and  since  the 
answer  is  obtained  by  dividing  the  20  apples  by  4,  it  follows  that  to 
divide  any  number  into  4  equal  parts,  that  is,  to  get,  one  fourth  ( J) 
of  a  number,  divide  it  by  4. 

In  like  manner,  to  divide  any  number  into  3  equal  parts,  that  is, 
to  get  one  third  of  a  number,  divide  it  by  3;  to  divide  any  number  into 
equal  parts,  that  is,  to  get  one  half  of  a  number,  divide  it  by  2,  -&c. 

6.  If  7  yards  of  cloth  cost  $35,  what  is  1  yard  worth? 

7.  A  school  of  60  pupils  is  to  be  divided  into  6  classes: 
how  many  will  there  be  in  each  class? 

8.  I  walked  54  miles  in  9  hours :  how  many  miles  per 
hour  did  I  go? 

EXAMPLES  FOR  PRACTICE. 

9.  Divide    75191  by    3  ....  Am.    250631 

10.  Divide  171654  by    4  .     ..."  Am.    4291 3| 

11.  Divide  512653  by    5  ....  Am.  102530* 

12.  Divide  534959  by    7  ....  Am.     >76422f 

13.  Divide  986028  by    8  ....  Am.  1232531 

14.  Divide  986974  by  11  .    .    .    .  Am.  89724J? 

In  the  following  examples,  find  the  continued  quotient  of  the 
given  number,  divided  by  2,  4,  6,  8,  10,  and  12;  that  is,  divide  the 
given  number  by  2,  then  the  resulting  quotient  by  4,  and  so  on. 

EXAMPLES.  AN8.  EXAMPLES.  AN8. 


15.  138240  ...     3. 

16.  230400  ...    5. 

17.  322560  .    .    .7. 

18.  506880  .         .  11. 


19.  783360  ...  17. 

20.  1336320     .    .  29. 

21.  1658880     .    .  36. 

22.  2488320         .  54. 


23.  At  $6  a  head,  how  many  sheep  can  be  bought  foi 
$222?  Am.  3*7. 

24.  At  $5  a  barrel,  how  many  barrels  of  flour  can  be 
bought  for  $895?  Am.  179. 

25.  If  8  pints  make  a  gallon,  how  many  gallons   are 
there  in  2520  pints?  Am.  315. 

REVIEW. — 62.  Give  an  example  of  the  first  class.  Explain  it.  Give 
an  example  of  the  second  class.  Explain  it.  How  is  the  half,  third,  «fcc- 
uf  any  number  obtained? 


40  RAY'S   HIGHER   ARITHMETIC. 

26.  At  the  rate  of  9  gallons  an  hour,  how  long  will  it 
Lake  to  fill  a  cistern  of  2115  gallons?    Ans.  235  hours. 

27.  What  number  multiplied  by  11  gives  638? 

Ans.  58. 

28.  If  three  acres  of  land  cost  $741,  what  is  the  cost 
of  1  acre?  Ans.  $247. 

29.  .If  $1715  be  divided  into  7  equal  parts,  how  many 
dollars  will  there  be  in  each  part?  Ans.  $245. 

30.  How   many  full  weeks   in    13    days?      How  many 
in  670  days?  Ans.  1  in  13,  and  95  in  670  days. 

31.  How  many  times  can  9  be  subtracted  from  155? 

Ans.  17  and  2  left. 

ART.  63.  Short  Division  is  performed  mentally,  and 
only  the  result  is  written;  but  when  all  the  work  is  writ- 
ten, it  is  called  Long  Division. 

Short  Division  is  generally  used  when  the  divisor  does 
not  exceed  12;  Long  Division,  when  the  divisor  exceeds 
12.  To  show  the  difference  between  them,  find  how  often 
5  is  contained  in  820. 

SHOET    DIVISION.  LONG    DIVISION. 

5)820  5)820(164  Quotient. 

164  Quotient. 

3  2  tens. 
Both  operations  are  performed  on        3  Q 

the  same  principle.       In  the  first,        

the  subtraction  is  performed  men-  ^  ^  units. 

f\  A 

tally;  in  the  second,  it  is  written. 

ART.  64.    RULE  FOR  LONG  DIVISION. 

1.  Draw  curved  lines  on  the  right  and  left  of  the  dividend, 
placing  the  divisor  on  the  left. 

2.  Find   how  often  the  divisor  is  contained  in  the   left-hand 
figure,  or  figures,  of  the  dividend,  and  write  the  number  in  the 
quotient  at  the  right  of  the  dividend.     . 

3.  Multiply  the  divisor  by  this  quotient  figure,  and  write  the 
product  under  that  part  of  the    dividend  from   which  it  was 
obtained. 


Re  VIEW. — 63.    Explain   the   difference   between   Short   Division    and 
Long  Division.     When  ia  the  latter  used  ? 


DIVISION   OF   SIMPLE   NUMBERS.  41 

4.  Subtract  this  product  from  the  figures  above  it ;   to  the  re- 
mainder bring  down  the  next  figure  of  the  dividend,  and  divide 
as  before,  until  all  the  figures  of  the  dividend  are  brought  down. 

5.  If  at  any  time  after  a  figure  is  brought  down,  the  number 
thus  formed  is  too  small  to  contain  the  divisor,  a  cipher  must  be 
placed  in  the  quotient,  and  another  figure  brought  down,  after 
which  divide  as  before. 

PROOF. — The  same  as  in  Short  Division. 

Divide  $4225  equally  among  13  men. 

SOLUTION. — As  13  is  not  contained  §^W-S    ^  oj -3 

in  4  (thousands),   therefore,    the  quo-  J  ja  J'l    j-  3  '§ 

tient  has  no  thousands.     Next,  take  42  13)4225(325 

(hundreds)  as  a  partial  dividend;  13  is  g  Q  nun 

contained   in   it  $    (hundreds)   times ;  — — 

after  multiplying  and  subtracting,  there  j"  ^ 

are  3  hundreds  left.     Then  bring  down  ^^_  ten8- 

2  tens,  and  32  tens  is  the  next  partial  6  5 

dividend.      In  this  13  is  contained  2  65  units, 
(tens)  times,  with  a  remainder  6  tens. 

Lastly,  bringing  down  the  6  units,  IS  is  contained  in  65  (units)  ex- 
actly 5  (units)  times.  The  entire  quotient  is  3  hundreds  2  tens  and 
5  units,  or  325. 

DEMONSTRATION. — This  operation  involves  no  principle  not  ex- 
plained in  Short  Division,  (Art.  61).  This  may  be  further  shown  by 
decomposing  the  dividend  into  parts,  each  exactly  divisible  by  13, 
as  follows : 

DIVISOR.  DIVIDEND.  QUOTIENT. 

13)  3900  +  260  +  65  (300  +  20  +  5 
3900 

+  260 
+  260 

+  65 

+  65 

NOTES. — 1.  The  product  must  never  be  greater  than  the  partial 
lividend  from  which  it  is  to  be  subtracted;  if  so,  the  quotient  figure 
is  too  large,  and  must  be  diminished. 

2.  The  remainder  after  each  subtraction   must  be  less  than  the 
iivisor;   if  not,  the  last  quotient  figure  is  too  small,  and  must  be 
increased. 

3.  The  order  of  each  quotient  figure  is  the  same  as  the  lowest 
jrder  in  the  partial  dividend  from  which  it  was  derived. 

4 


RAY'S   HIGHER   AIUT.rtlETIC. 


ART.  65.  In  the  Italian  method  of  Long  Division,  th« 
multiplication  and  subtraction  are  both  performed  in  the 
mind  at  the  same  time.  It  is  illustrated  thus : 

Divide  2087*7514  by  3256. 

SOLUTION.— The  first  figure  ITERATION.  Quot. 

it  the  quotient,  being  0,  mul-      3256j20877514(6412 
tiply  the    divisor   by  it,   and  13415 

lubtract  the  product  from  the  3911 

dividend,  saying  6  times  6  are  6554 

36,  and  as  36  can  not  be  taken  /j.  2  Rem 

from  7,  add  3  tens  or  30  to  the 

figure  of  the  dividend,  making  37,  and  then  36  from  37  leaves  1 ;  to 
compensate  for  the  3  tens  added  to  the  upper  figure,  carry  3  to  the 
next  product  when  it  is  formed,  saying  G  times  5  are  30  and  3  are  33, 
and  33  from  37  leaves  4,  there  being  3  to  carry :  then,  6  times  2  are 
12,  and  3  are  15,  and  15  from  18  leaves  3,  there  being  1  to  carry; 
6  times  3  are  18,  and  1  are  19,  and  19  from  20  leaves  1,  with  2  to 
carry,  and  2  from  2  leaves  0.  So  proceed  until  the  figures  of  the 
dividend  are  all  used. 


EXAMPLES  FOR   PRACTICE. 

1.  Divide  139180  by  453. 


PROOF. 

307  Quotient. 
453  Divisor. 


3280 
3171 

109 


921 
1535 

1228 

139071 

109  Rem. 


139180 


2. 
3. 
4. 
5. 

6. 
7. 
8. 
9. 
10. 


1004835  —  33  \.    .    .    .     =  30449i! 

5484888—67 =81864. 

4326422—961  =4502. 


1457924651  —  1204. 

65358547823  —  2789. 

33333333333—5299. 
245379683477  —  1263. 
555555555555—123456 
555555555555  —  654321 


.  =  1210900  Hro1} 
= 23434402 
.  =  629049552%% 
=  1942831611111 


=  849056111111 


DIVISION    OF   SIMPLE   NUMBERS.  43 

In  the  following,  multiply  A  by  itself,  also  B  by  itself:  divide  th« 
iifference  of  the  products  by  the  sum  of  A  and  B. 
A.  B. 

1902 Ans.   984. 

12.  8473    9437 Ans.    964. 

13.  2856    3765 Ans.    909. 

14.  33698   42856 Ans.   9158 

15.  47932   152604 Ans.  104672. 

In  the  following,  multiply  A  by  itself,  also  B  by  itself:  divide  th« 
difference  of  the  products  by  the  difference  of  A  and  B. 
A.  B. 

10.  4986  5369 Ans.  10355. 

17.  3973  4308 Ans.   8281. 

18.  23798  59635 Ans.  83433. 

19.  47329  65931 Am.  113260. 

20.  If  25  acres  produce  1825  bushels  of  wheat,  hov\ 
much  is  that  perWre?  ./Ins.  73  bushels. 

21.  How  many  times  1024  in  1048576?    Ans.  1024. 

22.  How  many  sacks,  each  containing  55  pounds,  can 
be  filled  by  2035  pounds  of  flour?  Ans.  37. 

23.  How  many  pages   in   a   book  of  7359   lines,  each 
page  containing  37  lines?  -4ns.  1983! 

24.  In  what  time  will  a  vat  of  10878  gallons  be  filled, 
at  the  rate  of  37  gallons  an  hour  ?         -4ns.  294  hours. 

25.  In  what  time  will  a  vat  of  3354  gallons  be  emptied, 
at  the  rate  of  43  gallons  an  hour?  Ans.  78  hours. 

26.  The  product  of  two  numbers  is  212492745;  one  is 
1035;  what  is  the  other?  Ans.  205307. 

27.  What  number  multiplied  by  109,  and  98  added  t« 
the  product,  will  give  106700  ?   "  Ans.  978. 

CONTRACTIONS  IN  DIVISION. 

<3ASE  I. — WHEN   THE  DIVISOR  IS  A  COMPOSITE  NUMBER. 

ART.  66.  RULK. — 1.  Divide  the  dividend  by  one  of  the  factor t 
of  the  divisor,  and  this  quotient  by  another  factor,  and  so  on,  until 
all  the  factors  have  been  used.  The  last  quotient  will  be  the  out 
required. 

REVIEW. — 64.  Give  the  rule ;  the  proof.  Explain  the  example.  When 
is  the  quotient  figure  too  large  ?  When  too  small  ?  How  is  the  order  of 
each  quotient  figure  known  7 


14  KAY'S   HIGHER   ARITHMETIC. 

2.  To  Jind  the  true  remainder :  Multiply  each  remainder  by  all 
he  preceding  divisors,  except  that  which  produced  it,  and  to  (ht 
mm  of  (he  products,  add  the  remainder  from  the  first  division 

Divide  21*7  by  15. 

SOLUTION.— Since  15  =  3X5,  divide      3)217 
by  3  and  then  by  5.     The  true  remainder         5)72  and  1  rem 

ifl  2  X  3  4- 1  =  7.     Hence,  the  quotient  is  i  A 

'         „  14  and  2  tern. 

14,  remainder  7. 

DEMONSTRATION. — Dividing  by  3  and  then  by  6,  is  the  same  as 
dividing  by  15;  for,  in  the  former  case,  the  quotient  must  be  multi- 
plied by  5  and  then  by  3,  and,  in  the  latter  case,  by  15,  to  produce 
the  given  dividend ;  and,  since  multiplying  by  5  and  then  by  3,  is 
the  same  as  multiplying  by  15,  (Art.  53),  the  quotients  on  which  these 
operations  are  performed  must  be  alike,  or  they  could  not  both  pro- 
duce the  same  dividend. 

To  prove  the  rule  for  finding  the  remainder;  dividing  217  by  3, 
the  quotient  is  72  threes,  and  1  unit  remainder.  Dividing  by  5,  the 
quotient  is  14  (fifteens),  and  a  remainder  of  2  threes.  Hence,  the 
whole  remainder  is  2  threes  -|-  1,  or  7. 

EXAMPLES   FOR   PRACTICE. 


3.  2332-=-54  .  = 

4.  3800^-56  .  =6 


1.  1036-=-28  .  =   37. 

2.  3640-^35   .  =104. 

5.  34855^-168    .     .    =   207T7g9s 

6.  620314^231    .     .    =26853Vi 

CASE    II.  —  WHEN    THE    DIVISOR    IS    ONE    WITH    CIPHERS 
ANNEXED,  AS  10,  100,   1000,  &C. 

ART.  67.  RULE. — Cut  off  as  many  figures  from  the  right  of 
the  dividend  as  there  are  ciphers  in  the  divisor;  the  figures  cut 
off  will  be  the  remainder,  the  other  figures  will  be  the  quotient. 

Divide  23543  by  100. 

1|00)235|43 

235  Quotient,  43  Remainder. 

ANALYSIS. — To  divide  23543  by  100,  is  to  find  how  many  hun- 
dreds it  contains.     This  is  done  by  mere  inspection,  the  part  on  the 
ight  of  the  line,  43  (units),  being  the  remainder,  since  it  is  less  than 
100:  while  that  on  the  left,  235,  is  the  quotient,  being  the  number  of 
hundreds  in  the  given  number. 

REVIEW. — 65.  What  is  the  Italian  method  of  long  division  ?  Explain 
the  example.  66.  What  is  the  rule  for  dividing  by  a  composite  number? 


CONTRACTIONS   IN   DIVISION.  45 

1.  4567—100 =   45-flfr 

2.  8325  —  1000 =     8-$fo 

3.  95043  —  100 =950TVo 

4.  730015—10000 =   73Ti)VW 

CASE  III. — WHEN  CIPHERS  ARE  ON  THE  RIGHT  OP  THE 
DIVISOR. 

ART.  68.    ROLE.— 1.   Out  off  the  ciphers  at  the  right  of  tht 
divisor,  and  as  many  figures  from  the  right  of  the  dividend. 

2.  Divide  the  remaining  part  of  the  dividend  by  the  remaining 
part  of  the  divisor. 

3.  Annex  the  figures  cut  off  to  the  remainder,  and  it  will  give 
the  true  remainder. 

Divide  3846  by  400. 

4|00)38|46 

9  Quotient,  20  0  +  46  =  246  Eem. 

DEMONSTRATION. — To  divide  by  400  is  the  same  as  to  divide 
by  100  and  then  by  4,  (Art.  66).  .  Dividing  by  100  gives  38,  and  46 
remainder,  (Art.  67) ;  then  dividing  by  4  gives  9,  and  2  remainder- 
the  true  remainder  is  2  X  100  +  46  =246,  (Art.  66.) 

EXAMPLES  FOR  PRACTICE. 

1.  34500  —  130 =  2651%  , 

2.  82500  —  1100 =75. 

3.  60000  —  1700 =  35TV& 

4.  1896000  —  24000 =79. 

CASE  IV. — WHEN  THE  DIVISOR  WANTS  BUT  LITTLE  OP 
BEINO  100,  1000,  10000,  &C. 

ART.  69.  RULE. — Cut  off  from  the  right  of  the  divi^snd,  by  a 
vertical  line,  as  many  figures  as  the  divisor  contains  ;  multiply  th« 
part  on  (he  left  of  the  line  by  what  the  divisor  wants  of  being  100, 
1000,  &c.,  and  set  the  product  under  the  dividend,  commencing 
at  units'  place.  Multiply  the  part  of  this  product  on  the  left  of  the 

REVIEW. — 66.  Explain  the  rule.  How  is  the  true  remainder  found? 
67.  What  is  the  rule  for  dividing  by  1  with  ciphers  annexed?  Explain 
the  example.  68.  What  is  the  rule  for  dividing  by  a  number  that  bM 
otpbers  on  its  right?  Explain  the  example. 


4t> 


RAY  8  HIGHER   ARITHMETIC. 


line  by  the  same  multiplier,  and  set  down  as  before.  Continue  so 
until  no  figure  falls  to  the  left  of  the  line;  then  add  the  several 
results,  and  for  every  1  carried  across  the  line,  add  to  the  sum 
already  obtained  on  the  right  of  the  line,  the  number  used  as  a 
multiplier.  This  will  be  the  true  remainder,  and  the  part  on  the 
left,  the  quotient. 

NOTE.  —  If  the  remainder  is  larger  than  the  divisor,  carry  one  to 
be  quotient,  and  the  excess  will  be  the  true  remainder. 

Divide  3289662  by  95. 

ANALYSIS.  —  The  reason  of  the  rule  can 
be  seen  in  the  explanation  of  this  example. 
The  divisor  is  assumed  to  be  100,  and  the 
quotient,  on  that  supposition,  is  32896,  with 
a  remainder  62.  If  this  quotient  were 
multiplied  by  the  true  divisor  95,  which  is 
6  less  than  the  one  assumed,  the  product, 
after  the  remainder  62  was  added  in,  would 
be  less  than  the  dividend  by  5  times  the 
quotient  32896.  Therefore,  32896  and  62 
are  the  true  quotient  and  remainder  for  only 
apart  of  the  dividend;  the  surplus  remain- 
ing to  be  divided  is  5  times  3289G  or  164480. 


OPERATION. 

95)32896162 
1644  80 
82  20 
10 

•JO 

92 
5% 

3462S'~ 
Qaot.     9  5 

Rem.    2 


34627 


This  surplus  is  divided  like  the  first  dividend,  the  quotient  and  re- 
mainder going  to  increase  those  already  obtained.  There  is  a  surplus 
also  in  this  operation,  which  is  treated  like  the  first,  and  so  on,  until 
the  quotient  is  nothing,  when  the  successive  divisions  must  cease. 

The  sum  of  the  remainders  is  192  ;  from  which,  after  setting  down 
02,  we  carry  1  to  the  first  column  of  the  quotients.  This  1  is  100, 
and  as  1  must  be  carried  to  the  quotient  for  every  95  in  the  remain- 
der, 5  out  of  this  100  should  continue  in  the  remainder,  and  be 
added  to  the  92  already  set  down,  making  97;  but  97,  being  large 
enough  to  contain  the  divisor  95,  must  furnish  another  unit  to  the 
quotient,  making  it  34628,  and  the  excess  2  is  the  true  remainder. 


EXAMPLES   FOR   PRACTICE. 


1.     75076822- 
2.    24206778- 
3.    680571293- 
4.   8662309749- 
5.  52351346504- 
6.  42104678535- 

-99.   . 
-989  . 
-9996  . 
-9994  . 
-99930 
-99800 

=  758351!$ 

=    2447  O^'i 


=  866751 

=  523880-^:M 

=  421890SSIB8 


FUNDAMENTAL   RULES  47 


GENERAL    PRINCIPLES   OF    MULTIPLICATION 
AND    DIVISION. 

ART.  70.  THEOREM  I.  —  Multiplying  either  factor  of  a 
product  multiplies  the  product  by  the  same  number. 

DEMONSTRATION. — 9  multiplied  by  5  gives  a  product  45,  and  9 
multiplied  by  10  (2  times  5)  gives  a  product  90,  which  is  2  times  45, 
and  so  in  all  cases;  for,  when  a  multiplier  is  made  2,  3,  4,  &c.  times 
as  large  as  ut  first,  the  multiplicand  must  be  taken  2,  3,  4,  &c.  times 
as  often  as  before,  and,  therefore  the  product  will  be  2,  3,  4,  &c.  times 
as  large  as  the  first  product.  And  since  this  is  true  of  the  multi- 
plier, it  is  also  true  of  the  multiplicand,  because  it  can  be  used  as 
the  multiplier.  (Art.  47.) 

ART.  71.  THEOREM  II.  —  Dividing  either  factor  of  a 
product  divides  the  product  by  the  same  number. 

DEMONSTRATION. — 9  multiplied  by  10  gives  a  product  90,  and 
9  multiplied  by  5  (  ^  of  10)  gives  a  product  45,  which  is  A  of  90,  and 
so  in  all  cases;  for  when  a  multiplier  is  made  A,  -3,  j,  &c.,  as  large 
as  at  first,  the  multiplicand  must  be  taken  ^,-3,  i>  &c->  as  often  as 
before,  and,  therefore,  the  product  will  be  5,  3>  |>  &c.,  as  large  as 
the  first  product.  And  since  this  is  true  of  the  multiplier,  it  is  also 
true  of  the  multiplicand,  because  it  can  be  used  as  the  multiplier. 
(Art.  47.) 

ART.  72.  THEOREM  III. — Multiplying  one  factor  of  a 
product,  and  dividing  the  other  factor  by  the  same  number, 
does  not  alter  the  product. 

DEMONSTRATION. — Multiplying  either  factor  by  2,  3,  4,  &c., 
multiplies  the  product  by  2,  3,  4,  &c.,  while  dividing  the  other  factor 
by  2,  3,  4,  &c.,  divides  the  product  by  2,  3,  4,  &c.  When  these  opera 
tions  are  performed  together,  the  product  is  both  multiplied  and  di- 
vided by  the  same  number,  and  must,  therefore,  remain  unchanged. 

ART.  73.  THEOREM  IV. — Multiplying  the  dividend,  or 
dividing  the  divisor,  by  any  number,  multiplies  the  quotient 
by  that  number. 

DEMONSTRATION. — If  any  divisor  is  contained  in  a  given  divi- 
dend a  certain  number  of  times,  it  must  be  contained  in  twice  that 
dividend,  twice  as  often ;  in  3  times  that  dividend,  3  times  as  often ; 


REVIEW. — 69.  What  is  the  rulo  for  dividing  by  a  number  that  want* 
but  little  of  being  100,  1000,  Ac.?  If  the  remainder  is  larger  than  the 
divisor,  what  tnuit  be  done?  Explain  example. 


48  RAY'S    HIGHER   ARITHMETIC. 


and  so  on.  Thus,  3  is  contained  in  12  four  times,  and  in  twice  12  it 
is  contained  twice  4,  =  8  times.  Again,  in  any  given  dividend,  half 
the  divisor  will  be  contained  twice  as  often  as  the  whole  divisor ;  one 
third  of  the  divisor,  3  times  as  often,  and  so  on ;  thus,  4  is  contained 
in  12  three  times,  and  one  half  of  4,  which  is  2,  is  contained  in  12, 
twice  3,  =  6  times. 

ART.  74.  THEOREM  V. — Dividing  the  dividend,  or  mul 
tiplying  the  divisor  by  any  number,  divides  the  quotient  by 
thai  number. 

DEMONSTRATION. — If  any  divisor  is  contained  in  a  given  divi- 
dend a  certain  number  of  times,  it  must  be  contained  in  half  that 
dividend,  half  as  many  times;  in  one  third  of  that  dividend,  one  third 
as  many  times,  and  so  on.  Thus,  3  is  contained  in  12  four  times, 
and  in  one  half  of  12,  which  is  six,  it  is  contained  one  half  of 
4  =  2  times. 

Again,  in  any  given  dividend,  twice  the  divisor  will  be  contained 
half  as  often  as  the  divisor  itself;  three  times  the  divisor,  one  third  as 
often,  and  so  on  ;  thus,  3  is  contained  in  24  eight  times,  and  2  times 
3,  which  is  6,  is  contained  in  24  one  half  of  8  =  4  times. 

ART.  75.  THEOREM  VI. — Multiplying  or  dividing  both 
dividend  and  divisor  by  the  same  number,  does  not  change 
the  quotient. 

DEMONSTRATION. • — This  follows  from  the  two  preceding  theorems; 
for  the  effects  produced  by  multiplying  or  dividing  both  dividend 
and  divisor  exactly  balance  each  other ;  thus,  6  is  contained  in  24 
4  times;  and  twice  6  is  contained  in  twice  24,  just  4  times :  also,  one 
half  of  6  is  contained  in  one  half  of  24,  just  4  times ;  and  so  of  any 
other  dividend  and  divisor. 

CONTRACTIONS    IN   MULTIPLICATION   AND 
DIVISION. 

ART.  76.  To  multiply  by  any  simple  part  of  100, 
1000,  &c., 

RULE. — Multiply  by  100,  1000,  &c.,  and  take  such  a  part  n 
the  result  as  the  multiplier  is  of  100,  1000,  &c. 

NOTE. — To  get  two  thirds  (f)  of  a  number,  divide  it  by  3,  which 
gives  one  third  (-3)  of  it,  (Art.  62),  and  multiply  the  quotient  by  'J ; 
or,  if  it  is  more  convenient,  multiply  the  number  by  2  first,  and 
divide  the  product  by  3.  In  like  manner,  to  get  three  fourths  ( |) 
of  a  number,  multiply  it  by  3,  and  divide  the  product  by  4,  and  so  ni. 


CONTRACTIONS  IN   MULTIPLICATION.  48 


PARTS  OF  100. 

PASTS  or  1000. 

12£  =  |  of  100. 

125  =  $ 

of  1000. 

IGJj  =  £  of  100. 

166§  =  ^ 

of  1000. 

25  =£oflOO. 

250  =  £ 

of  1000. 

33£  =  £  of  100. 

333J  =  J 

of  1000. 

37*  =  £  of  100. 

375  =g 

of  1000. 

62i  =  |  of  100. 

625  =  f 

of  1000. 

66f  =  §  of  100. 

666f  =  | 

of  1000. 

75  =  g  of  100. 

750  =f 

of  1000. 

83£  =  |  of  100. 

833$  =  g 

of  1000. 

87*  =  |  of  100. 

875  =| 

of  1000. 

Multiply  246  by  87i  24600 

SOLUTION.  —  Since  87%  is  |  of  100,  annex  two  7 

ciphers  to  the  multiplicand,  which  multiplies  it  g  ")!  7O  °0~0 

by  100,  (Art.  64),  and  then  take  g  of  the  result. 

(See  Note).  4ns.  21o25 

EXAMPLES   FOR  PRACTICE. 

1.  422x331    .    .    .     =      140661 

2.  3708x25      .    .    .     «=      92700. 

3.  6564x621    .    .    .     =    410250. 

4.  10724x161     .    .    .     =   178733s 

5.  8740x75      .    .    .     =«=    655500. 

6.  53840x125    .    .    .     =6730000. 

7.  4456X6665.    .    .     =29706661 

8.  7293X833}  .    .    .     =6077500. 

9.  852x875    .    .    .     =    745500. 
10.        64082x375    .    .    .*=  24030750. 

ART.  77.  To  multiply  by  any  number  whose  digits  are 
all  alike, 

RULE.  —  Multiply  as  if  the  digits  vxrt  9'*,  (Art.  50),  and  tak  .* 
nch  a  part  of  the  product  as  the  digit  w  of  9. 

Multiply  592643  by  66666. 

OOLCTION.  —  Multiply  6012G-J3  by  90990,  (Art.  56),  the  product  if 
5920.1707357;  take  |  of  this  product,  since  6  is  J  of  9,  the  result  is 
89009138238. 

REVIEW  .  —  70-75.  State  and  prove  the  theorem*. 
5 


M)  RAY'S   HIGHER   ARITHMETIC. 

1.  451402x3333     .....     =1504522806. 

2.  281257X555555      .    .    .     =  156:»537:h!035. 

3.  030224x4444000   .    .    .  =2800715450000. 

ART.  78.    To  divide  by  a  number  ending  in  any  simple 
part  of  100,  1000,  &c., 


Multiply  both  dividend  and  divisor  by  such  a  number 
3,  4,  o,  or  8,)  a*  will  convert  (he  final  figures  of  the  divisor  into 
iphers,  and  then  divide  the  former  product  by  the  latter. 

NOTE.  —  1.  If  there  be  a  remainder,  it  should  be  divided  by  the 
multiplier,  to  get  the  true  remainder. 

2.  The  multiplier  is  3,  4,  6,  or  8,  according  as  the  final  portion  of 
the  divisor  is  thirds,  fourths,  sixths,  or  eighth*  of  100,  1000. 

3.  Several  successive  multiplications  may  sometimes  be  made  be- 
fore dividing. 

Divide  6903141128  by  21875.    Am.  3155722Vs!75 

SOLUTION.  —  Multiply  both  by  8  and  4  successively.  The  divisor 
becomes  700000,  and  the  dividend  22090051G006,  while  the  quotient 
remains  the  same,  (Art.  75).  Performing  the  division  as  in  Art.  68, 
the  quotient  is  315572,  and  remainder  116096.  The  remainder  being 
a  part  of  the  dividend,  has  been  made  too  large  by  the  multiplication 
by  8  and  4,  and  is,  therefore,  reduced  to  its  true  dimensions  by 
dividing  by  8  and  4.  This  gives  3628  for  the  true  remainder. 

EXAMPLES  FOR  PRACTICE. 

1.  300521761  —  225    .    .    .     =    1335052-Vs 

2.  1510337204—43750    .    .     =     34521^.^ 

3.  22500712361  —  1400250     .     =  10000  rYoVsVo- 

4.  620712480—  208S3I     =  29794,  Rcm.  4140  J 

5.  742851692—29101       =254092   ..        25i 

ART.  79.  To  divide  by  any  number  uhosc  digits  arc 
all  alike, 

RULE.  —  Treat  the  dividend  a:id  divisor  alike  by  muUiplicaticn, 
end  division,  if  necessary,  until  the  dir/its  of  the  divisor  are  D's; 
then  divide  by  Rule  in  Art.  69.  _ 

REVIEW.  —  76.  What  is  the  rule  for  multiplying  by  any  simple  part  of 
100,  1000,  Ac.?  77.  For  multiplying  by  any  number  whose  dibits  nre  all 
alike?  78.  For  dividing  by  any  nombor  ending  in  any  simple  part  of 
180.  1000.  Ac? 


SUMMARY  OF  PRINCIPLES. 


NOTE.  —  1.  If  tho  divisor  ends  in  ciphers,  divide  first  as  in  Art.  68. 
2.  If  a  remainder  occurs,  reverse  upon  it  the  operations  performed 
on  the  dividend  and  divisor,  to  get  the  trut  remainder. 

Divide  420565342  by  666. 

SOLUTION.  —  Divide  both  numbers  by  2,  and  multiply  the  results 
by  3  ;  this  makes  the  divisor  999,  and  does  not  alter  the  quotient, 
(Art.  75),  which  is  found,  (by  Art.  69),  to  be  G31479,  with  a  remain- 
der 492  ;  the  latter  being  a  part  of  tha  dividend,  differs  from  th« 
true  remainder,  by  reason  of  the  division  by  2,  and  the  multiplication 
.by  3,  and  is  brought  to  its  true  dimensions  by  multiplying  by  2  and 
dividing  by  3. 

EXAMPLES  FOR  PRACTICE. 

1.  13641096—2222    .    .    .    .=        6139M* 

2.  376802902  —  7770    .    .    .    .=      484944rfo 

3.  8811857528—3333    .    .    .    .  —26488211  H4 

4.  5606258492  —  88888.    .    .    .=    630715VA 


ART.  80.  Notation  and  Numeration,  Addition,  Sul  • 
traction,  Multiplication,  and  Division,  are  called  th<*  fuii- 
damental  rules  of  Arithmetic,  because  all  the  vsrioun 
operations  of  Arithmetic  are  performed  by  them. 

ART.  81.  NOTATION  is  the  method  of  expressing  num- 
bers in  figures.  NUMERATION  is  the  method  of  expressing 
numbers  in  wordt. 

ART.  82.  ADDITION  is  the  process  of  finding  the  aggre- 
gate or  sum  of  two  or  more  numbers,  (Art.  40). 

ART.  83.  SUBTRACTION  is  the  process  of  finding  the 
difference  between  two  numbers,  (Art.  43). 

ART.  84.  MULTIPLICATION  is  the  process  of  finding 
what  is  produced  by  taking  one  number  as  many  timet  as 
there  are  units  in  another,  (Art.  46). 

REVIEW. — 79.  What  is  the  rule  for  dividing  by  a  number  whose  digits 
are  all  alike?  How  is  the  true  remainder  obtained?  80.  Which  are  th« 
fundamental  rules  of  Arithmetic?  Why  are  they  so  called?  81.  What  ii 
Notation?  Numeration?  82.  Addition?  83.  Subtraction? 


52  RAY'S   HIGHER   ARITHMETIC. 

ART.  83.    DIVISION  is  the  process  of  finding  how  many 
times  one  number  is  contained  in  another,  (Art.  53). 

GENEJIAL  PROBLEMS. 

1.  When   the  separate  cost  of  several  things  is  given, 
how  is  the  entire  cost  found? 

2.  When  the  sum  of  two  numbers  and  one  of  them  are 
given,  how  is  the  other 'found? 

3.  When    the  less   of  *.wo  numbers  and  the  difference 
between  them  arc  given,  how  is  the  greater  found? 

4.  When  the  greater  of  two  numbers  and  the  difference 
between  them  are  given,  how  is  the  less  found? 

5.  When  the  cost  of  one  article  is  given,  how  do  you 
find  the  cost  of  any  number  at  the  same  price? 

6.  When   the  divisor  and   quotient  are  given,  how  do 
you  find  the  dividend? 

7.  How  do  you  divide  a  number  into  parts  each  con- 
taining a  certain  number  of  units? 

8.  How  do  you  divide  a  number  into  a  given  number 
of  equal  parts? 

9.  If  the  product  of  two  numbers  and  one  of  them  is 
*ivcn,  how  do  you  find  the  other? 

10.  If  the  dividend  and  quotient  arc  given,  how  do  you 
find  the  divisor? 

11.  If  you  have  the  product  of  three  numbers,  and  two 
of  them  arc  given,  how  do  find  the  third  ? 

12.  If  tlie  divisor,  quotient,  and  remainder  arc  given, 
how  do  you  find  the  dividend? 

13.  If  the  dividend,  quotient,  and  remainder  arc  given, 
how  do  you  find  the  divisor? 

ART.  86.      PARTICULAR  EXAMPLES. 

1.  I  bought  3  horses  for  $165;   two  cows  for  $87; 
and  7  sheep  for  $29:  what  did  all  cost?       Ant.  $231. 

2.  The  sura  of  two  numbers  is  664,  and  one  of  them 
IB  369:  what  is  the  other?  Ans.  295. 

3.  The  difference  between  two  numbers  is  168,  and  the 
lees  number  is  289:  what  is  the  greater?         Ans.  457. 

REVIEW.— 84.  What  is  Multiplication  ?    85.  Division? 


SUMMARY  OF  PRINCIPLES.  53 

4.  The   greater  of  two  numbers  is  753,  and  their  dif- 
ference 457:  what  is  the  less?  Ant.  296. 

5.  At  $7  a  yard,  what  will  5  yards  of  cloth  cost? 
SOLUTION. —  6  yards  are  6  times  1  yard.     If  1  yard  cost  $7,  5 

yards  will  cost  5  times  $7,  which  are  $35,  Ana. 

G.  The  divisor  is  753,  and  the  quotient  245:   what  ii 
llie  dividend?  Ans.  184485. 

7.  How  many  classes  of  9  pupils  can  be  formed  in  a 
school  containing  54  pupils? 

SOLCTIO.V. — Since  it  takes  9  pupils  to  form  one  class,  there  will  be 
as  many  classes  as  9  pupils  are  contained  in  54  pupils;  9  in  54,  6 
times.  AM.  6. 

8.  If  you  divide  28  apples  equally  among  4  boys,  what 
will  be  the  share  of  each? 

SOLUTIO.V.  —  To  give  each  boy  1  apple,  will  require  4  apples; 
hence,  each  boy  will  receive  1  apple  as  often  as  4  apples  are  con- 
tained in  28  apples ;  4  in  28,  7  times.  Ans.  1  apples. 

9.  The  product  of  two  numbers  is  612451,  and  one  of 
them  is  203:  what  is  the  other?  Ans.  3017. 

10.  The    dividend    is    395631145    and    the    quotient 
4007:  what  is  the  divisor?  Ans.  98735. 

11.  The  product  of  three  numbers  is  195318005  ;  two 
are  307  and  703:  what  is  the  third?  Ant.  905. 

12.  What  number,  divided  by  473,  will  give  the  quo- 
ticlit  8061  and  remainder  365?  Am.  381321$. 

13.  The  dividend  is  7781174,  the  quotient  8216,  the 
remainder  622:  what  is  tho  divisor?  Ans.  947. 

ART.  87.    MISCELLANEOUS  EXERCISES. 

1.  A  grocer  gave  153  barrels  of  flous,  worth  $6   a 
barrel,  for  54  barrels  of  sugar:  what  did  the  sugar  cost 
per  barrel?  Ans.  $17. 

2.  When  the  divisor  is  35,  quotient  217,  and  remain- 
der 25,  what  is  the  dividend?  A'm.  7620. 

3.  What  number  besides  41  will  divide  4879  without 
a  remainder?  Ant.  119. 

4.  Of  what  number  is  103  both  divisor  and  quotient? 

Ans.  10609. 

5.  What  is  the  nearest  number  to  53815,  that  can  be 
divided  by  375  without  a  remainder?  -4ns.  54000. 


54  RAY'S   HIGHER   ARITHMETIC. 

•  6.  A  farmer  bought  25  acres  of  land  for  $2675:  what 
did  19  acres  of  it  cost?  Am.  $2033. 

7.  Bought  15  horses  at  $75  a  head:  at  how  much  per 
head  must  I  sell  them  to  gain  $210?  Ans.  $81). 

8.  A  locomotive  has  391  miles  to  run  in   11    hours: 
after  running  139  miles  in  4  hours,  at  what  rate  per  hour 
must  the  remaining  distance  be  run?         Ans.  36  miles. 

9.  A  merchant  bought  235  yards  of  cloth   at  $5   per 
yard:    after   reserving   12    yards,    what   will    he  gain   by 
selling  the  remainder  at  $7  per  yard?  -4ns.  $386. 

10.  A  grocer  bought  135  barrels   of  pork  for  $2295; 
he   sold   83   barrels   at  the  same  rate  at  which   he  pur- 
chased, and    the    remainder    at    an   advance    of    $2    per 
barrel:  how  much  did  he  gain?  Ans.  $104. 

11.  A  drover  bought  5  horses  at  $75  each,  and  12  at 
$68  each;   he  sold  them  all  at  $73   each:  what  did   he 
gain?  Ans.  $50. 

At  what   price  per  head   must   he  have  sold  them    to 
have  gained  $118?  Ans.  $77. 

12.  A    merchant   bought   3    pieces   of  cloth    of    equal 
lengths  at  $4  a  yard;  he   gained  $24  on  the  whole,  by 
selling  2  pieces  for  $240:  how  many  yards  were  there  in 
each  piece?  Ans.  18. 

13.  If  18  men  can  do  a  piece  of  work  in  15  days,  in 
how  many  days  will  one  man  do  it? 

SOLUTION. — It  will  require  1  man  18  times  as  long  as  18  rften. 
Eighteen  times  15  days  are  270  days.     Ans. 

14.  If  13   men  can   build  a  wall   in  15  days,  in  how 
many  days  can  it  be  done  if  8  men  leave?        Ans.  39. 

15.  If  14  men  can  perform  a  job  of  work  in  24  days, 
in  how  many  days  can  they   perform  it  with  the  assist- 
ance of  7  more  men?  Ans.  16  days. 

16.  A  company  of  45  men  have  provisions  for  30  days: 
how  many  men  must  depart,  that  the  provisions  may  last 
the  remainder  50  days?  Ans.  18  men. 

17.  A  horse  worth  $85,  and  3  cows  at  $18  each,  were 
exchanged  for  14  sheep  and  $41  in  money:  at  how  much 
each  were  the  sheep  valued?  Ans.  £7. 

18.  A  drover  bought  an   equal    number  of  sheep   an<J 
hogs  for  $1482;  he   gave   $7  for  a  sheep,  and  $6  lor  a 
hog:  what  number  of  each  did  he  buy?  An*.  114. 

SUGGESTION. — 1  sheep  and  1  hog  cost  $7  +  $G  =  $13. 


PROPERTIES  OF  NUMBERS.  55 

19.  A   trader    bought   a    lot   of  torses   and    oxen    for 
$1200;  the  horses  cost  $50,  and  the  oxen  $17  a  head; 
there  were  twice  as  many  oxen  as  horses:  how  many  were 
there  of  each?  Ans.  15  horses  and  30  oxen. 

20.  In   a  lot  of  silver  change  worth   1050  cents,  one- 
Beventh  of  the  value  is  in  25  cent  pieces;  the  rest  is  made 
up  of  10  cent,  5  cent,  and  3  cent  pieces,  of  each  an  equal 
number:  how  many  of  each  coin  are  there? 

Ans.  of  25  cent  pieces,  6;  of  the  others,  50  each. 

21.  A  speculator  had  140  acres  of  land,  which  he  might 
have  sold  at  $210  an  acre,  and  gained  $6300,  but  after 
holding,  he  sold  at  a  loss  of  $5000:  how  much  an  acre 
did  it  cost  him,  and  how  much  an  acre  did  he  sell  it  for? 

Ans.  $165,  cost;  and  $125,  sold  for. 


VII.  PROPERTIES  OF  NUMBERS. 

ART.  88.     DEFINITIONS. 

1.  An  integer  is  a  whole  number;  as,  1,  2,  3,  &c. 

2.  Whole  numbers  are  divided  into  two  classes — prime 
numbers  and  composite  numbers. 

3.  A  prime  number  is  one  that  can  be  exactly  divided 
by  no  other  whole  number  but  itself  and  unity,  (1);  as, 
1,  2,  3,  5,  7,  11,  &c. 

4.  A  composite  number   is    one    that    can   be   exactly 
divided  by  some  other  whole  number  besides  itself  and 
unity;  as,  4,  6,  8,  9,  10,  &o. 

REMARK. — Every  composite  number  is  the  product  of  two  or 
more  other  numbers. 

5.  Two  numbers  are  prime  to  each  other,  when  unity  is 
the  only  number  that  will  exactly  divide  both ;  as,  4  and  5. 

REMARK. — Two  prime  numbers  are  always  prime  to  each  other: 
•onetimes,  also,  two  composite  numbers:  as,  4  and  9. 

6.  An  even  number  is  one  whicu  can  be  divided  by  2 
rithout  a  remainder ;  as,  2,  4,  6,  8,  &c. 

R  E  v  t  E  w. — 83.  What  is  an  integer  T  Into  what  classes  are  integer! 
divi.1,',1?  What  is  a  prime  number?  a  composite  number?  When  are 
numbers  prime  to  each  other?  What  kind  of  numbers  must  be  prime  U 
each  other  ?  What  kind  may  be  ?  What  is  an  even  number  f 


5(5  HATS   HIGHER   ARITHMETIC. 

7.  An  odd  number  is  one  which  can  not  bo  divided  bj 
2  without  a  remainder;  as,  1,  3,  5,  7,  &c. 

RKMARK. — All  even  numbers  except  2  nre  composite:  the  odd 
numbers  arc  partly  prime  and  partly  composite. 

.  8.  A  divisor  or  measure  of  a  number,  is  a  number  (hat 
will  divide  it  without  a  remaindur:  2  is  a  divisor  of  4;  5 
of  10,  &c. 

9.  One  number  is  divisible  by  another  when  it  contains 
that  other  without  a  remainder;   8  is  divisible  by  2. 

10.  A  multiple  of  a  number  is  the  product  obtained  by 
taking  it  a  certain  number  of  times;  15  is  a  multiple  of 
5,  being  equal  to  5  taken  3  times ;  hence, 

1st.  A  multiple  of  a  number  can  always  be  divided  by 
it  icithout  a  remainder. 

2d.  Every  multiple  is  a  composite  number. 

11.  Since  every  composite  number   is   the   product  of 
factors,  (Def.  4),  each  factor  must  divide  it  exactly;  hence, 
every  factor  of  a  number  is  a  divisor  of  it. 

12.  A  jyrime  factor  of  a    number   is   a   prime    number 
that  will  exactly  divide   it:    5   is   a  prime   factor  of  20; 
while  4  is  a  factor  of  20,  not  a  prime  factor;  hence, 

1st.  The  prime  factors  of  a  number  are  all  the  prime 
numbers  that  will  exactly  divide  it;  1,  2,  3,  and  5,  are 
the  prime  factors  of  30. 

2d.  Every  composite  number  is  equal  to  the  product 
of  all  its  prime  factors.  All  the  prime  factors  of  15  are 
1,  3,  and  5  ;  and  1  X  3  X  5  =  15. 

13.  Any  factor  of  a  number  is  called  an  aliquot  part  of 
it;  1,  2,  3,  4,  and  6,  arc  aliquot  parts  of  12. 


FACTORING. 

ART.  89.  Factoring  depends  on  the  following  PRIN- 
CIPLES and  PROPOSITIONS. 

PIUNCIPI.K  1.  A  factor  of  a  number  is  a  factor  of  any 
multiple  nf  lluit  number. 

DEMONSTRATION. — Since  G  =  2X3,  therefore,  any  multiple  of 
6  =  '2  X  3  X  some  number ;  hence,  every  factor  of  0  is  also  a  facto? 
tf  the  multiple. 


FACTORING. 


PRINCIPLE  2.  A  factor  of  any  two  numbers  is  also  a 
factor  of  their  sum. 

,    DEMONSTRATION. —  Since  each  of  the  numbers  contains   (lie 
factor  a  certain   number  of  times,  their  sura   must   contain    it    as 
often  as  both  tlie  numbers;  2,  which  is  a  factor  of  G  ninl  10,  nuibi 
be  a  factor  of  their  sum,  for  6  is  3  twos,  and  10  is  5  twos,  and  thei 
Mm  is  3  twos  +  5  twos  =  8  twos. 

A.RT.  90.    From  these  principles  are  derived  six 

PROPOSITIONS. 

PROP.  I. — Every  number  ending  with  0,  2,  4,  G,  or  8,  is 
divisible  ly  2. 

DEMONSTRATION. — Every  number  ending  with  a  0,  is  either  10 
or  some  number  of  tens ;  and  since  10  is  divisible  by  2,  therefore,  by 
Principle  1st,  (Art.  80),  any  number  of  tens  is  divisible  by  2. 

Again,  any  number  ending  with  2,  4,  G,  or  8,  may  be  considered 
as  a  certain  number  of  tens  plus  the  figure  in  the  units'  place;  and 
since  each  of  the  two  parts  of  the  number  is  divisible  by  2,  therefore, 
by  Principle  2d,  (Art.  89),  the  number  itself  is  divisible  by  2;  thus, 
86  =  30-1-6  =  3  tens -f- 6;  each  part  is  divisible  by  2,  hence,  3C  is 
divisible  by  2. 

Conversely,  no  number  is  divisible  ly  2,  unless  it  ends 
with  0,  2,  4,  6,  or  8. 

PROP.  II. — A  number  is  divisible  by  4,  when  the  number 
denoted  by  its  two  right  hand  digits,  is  divisible  by  4. 

DEMONSTRATION. — Since  100  is  divisible  by  4,  any  number  of 
hundreds  will  be  divisible  by  4,  (Art.  80,  Principle  1st);  and  any 
number  consisting  of  more  than  two  places  may  be  regarded  as  a 
certain  number  of  hundreds  plus  the  number  expressed  by  the  digits 
in  tens'  and  units'  places,  (thus,  381  is  equal  to  3  hundreds-}-  81) ; 
then,  if  (he  latter  part  (8-J)  is  divisible  by  4,  both  parts,  or  the 
number  itself,  will  be  divisible  by  4,  (Art.  80,  Prin.  2d). 

Conversely,  no  number  is  divisible  by  4,  unless  the  number 
dctiotcd  by  its  tico  right-hand  digits  is  divisible  by  4. 

R  E  vi  K  w. — 88.  What  is  nn  odd  number?  Arc  the  even  numbers  prime 
or  composite?  Are  odd  numbers  prime  or  composite?  What  is  a  divisor 
•)f  a  nmnWr?  When  if  one  nuiul>cr  divi*il>le  b.v  another  ?  What  iy  n 
multiple  of  11  nuiubor?  What  two  :i.\i'.m.-  cunporninj;  multiples?  Wluii  u 
a  factor  of  a  nuinWr?  A  priiuo  factor?  What  two  axioms  concerning 
prime  factors?  What  u  aa  aliquot  part  of  a  numocr?  Give  example! 
illustrating  the  definitions. 


S8  RAY'S   HIGHER  ARITHMETIC. 

PROP.  III. — A  number  ending  in  0  or  5  is  divisible  ly  5. 

DEMONSTRATION. — Ten  is  divisible  by  6,  and  every  number  jf 
two  or  more  figures,  is  a  certain  number  of  tens,  plus  the  right  hand 
digit;  if  this  is  5,  both  parts  of  the  number  are  divisible  by  6,  and, 
hence,  the  number  itself  is  divisible  by  5,  (Art.  89,  Prin.  2d). 

Conversely,  no  number  is  divisible  by  5,  unless  it  ends  in 
0  or  5. 

PROP.  IV. — Every  number  ending  in  0,  00,  &c.,  is  dl 
visible  by  10,  100,  &c. 

DEMONSTRATION. — If  the  number  ends  in  0,  it  is  either  10  or  a 
multiple  of  10;  if  it  ends  in  00,  it  is  either  100,  or  a  multiple  of  100, 
and  so  on ;  hence,  by  Prin.  1st,  Art.  89,  the  proposition  is  true. 

PROP.  V. — A  composite  number  is  divisible  by  the  product 
of  any  two  or  more  of  its  prime  factors. 

DEMONSTRATION. — Since  2  X  3X5  =  30,  it  follows  that  2X3 
taken  6  times,  makes  30;  hence,  30  contains  2  X  3  (6)  exactly  6  times. 
In  like  manner,  30  contains  3X5  (15)  exactly  2  times,  and  2X5 
(10),  exactly  3  times. 

Hence,  if  any  even  number  is  divisible  by  3,  it  is  also 
divisible  by  G. 

DEMONSTRATION. — An  even  number  is  divisible  by  2;  and  if 
also  by  3,  it  must  be  divisible  by  their  product  2  X  3,  or  6. 

PROP.  VI. — Every  prime  number,  except  2  and  5,  end* 
with  1,  3,  V,  or  9. 
DEMONSTRATION. — This  is  in  consequence  of  Prop.  I  and  III. 

ART.  91.  To  find  the  prime  factors  of  a  composite 
number, 

RULE. —  Dimde  the  given  number  by  any  prime  number  that 
tcill  exactly  divide  it;  divide  the  quotient  in  like  manner,  and  so 
continue  until  the  quotient  is  a  prime  number;  the  last  quotient 
and  the  several  divisors  are  the  prime  factors. 

REVIEW. — 89.  What  is  the  first  principle  used  in  factoring  ?  Prove  it 
What  is  the  second  principle  used  in  factoring?  Prove  it.  90.  When 
is  a  number  divisible  by  2,  and  when  not?  Why?  When  by  4,  and  when 
net?  Why  ?  When  by  5,  and  when  not  ?  Why  ?  When  by  10,  100,  Ac., 
md  when  not?  Why?  What  is  every  composite  number  divisible  by? 
Why?  If  an  even  number  is  divisible  by  3,  what  else  must  divide  it? 
Why?  How  do  the  prime  numbers  end?  Why  ?  91.  What  is  the  rule  fc» 
the  prime  factors  of  a  number? 


FACTORING.  59 


REMARKS. — 1.  Divide  first  by  the  smallest  prime  factor. 

2.  The  least  divisor  of  any  number  is  a  prime  number;  for,  if  it 
were  a  composite  number,  its   factors,  "which  are  less  than  itself, 
would  also  be  divisors,  (Art.  89),  and  then  it  would  not  be  the 
least  divisor.     Therefore,  the  prime  factors  of  any  number  may  be 
found  by  dividing  it  first  by  the  least  number  that  will  exactly  divide 
it,  then  dividing  this  quotient  in  like  manner,  and  so  on. 

3.  Since  1  is  a  factor  of  every  number  either  prime  or  composite... 
il  is  not  usually  specified  as  a  factor. 

Find  the  prime  factors  of  42. 

DEMONSTRATION. — By  trial,2  is  found  to  be  a  factor     V\AO 

of  42.    Also,  3  is  found  by  trial  to  be  a  factor  of  21,  and     — ^ 

consequently  a  factor  of  42,  which  is  a  multiple  of  21,      3)21 
(Art.  89).     In  like  manner,  7  being  a  factor  of  21,  must  >j 

be  a  factor  of  42 ;  and  since  2  X  3  X  7  =  42,  there  can 
be  no  other  factors  of  42  besides  2,  3,  and  7. 

SEPARATE    INTO    PRIME    FACTORS, 


-1.  45  .    .  Ans.  3,3,5. 

2.  48  Ana.  2,2,2,2,3. 

3.  50  .     .  Ans.  2,5,5. 

4.  54  .   Ans.  2,3,3,3. 

5.  56  .    Ans.  2,2,2,7. 


6.  60    .   ^ns.  2,2,3,5. 

7.  63   .    .  Ans.  3,3,7. 


8.  72  Ans.  2,2,2,3,3. 

9.  75  .     .  Am.  3,5,5. 

10.  80  Ans.  2,2,2,2,5. 

11.  84  .    Ans.  2,2,3,7. 

12.  96  Ans.  2,2,2,2,2,3. 

13.  98  .    .  Am.  2,7,7. 

14.  99  .     Am.  3,3,11. 


THE  PCPIL  who  desires  to  be  an  expert  Arithmetician,  should  be 
able  to  give  the  prime  factors  of  all  numbers  under  100  by  mere 
inspection,  the  operation  being  performed  mentally. 

15.  Factor  210 Ans.  2,  3,  5,  7. 

16.  Factor  1155 Ans.  3,  5,  7,  11. 

17.  Factor  10010 Ans.  2,  5,  7,  11,  13. 

18.  Factor  36414.    .     .     .     Ans.  2,  3,  3,  7,  17,  17. 

19.  Factor  58425 Ans.  3,  5,  5,  19,  41. 

The  prime  factors  common  to  several  numbers  may  be 
found  by  resolving  each  into  its  prime  factors,  then  taking 
the  prime  factors  alike  in  all. 

REVIEW. —  91.  Which  number  is  it  most  convenient  to  divide  by  first  1 
What  is  said  of  the  least  divisor  of  a  number?  Of  1,  as  a  factor?  Explain 
(he  example,  and  show  the  reason  of  the  rule. 


60  RAY'S  HIGHER   ARITHMETIC. 


Find  the  prime  factors  common  to 

20.  42     and  98 Am.  2,  7. 

21.  45     and  105 AM.  3,  5. 

22.  90     and  210    . AM.  2,  3,  5. 

23.  210  and  315 AM.  3,  5,  7. 

ART.  92.  Since  any  composite  number  is  divisible,  nol 
only  by  each  of  its  prime  factors,  but  also  by  the  product 
of  any  two  or  more  of  them,  (Art.  90,  Prop.  V.);  hence, 

To  find  all  the  divisors  of  any  composite  number, 

Jlri.E. — Resolve  the  number  into  its  prime  factors;  and  then 
form  from  these  factors  all  the  different  products  of  which  they 
will  admit;  the  prime  factors  and  their  products  will  be  all  tlit 
dirisors  of  the  given  number. 

42  =  2  X  3  X  7;  and  all  its  divisors  are  2,  3,  7,  and 
2  X  3,  2  X  7,  and  3  X  7;  or,  2,  3,  7,  6,  14,  21. 

Find  all  the  divisors 

1.  Of  70       .     .  Ans.  2,  5,    7  and  10,  14  and  35". 

2.  Of  30       .     .  Ans.  2,  3,     5  and    6,  10  and  15. 

3.  Of  196    .     .  Am.  2,  7,    4,  14,  28,  49  and  98. 

4.  Of  231    .     .  Ans.  3,  7,  11  and  21,  33  and  77. 


GREATEST   COMMON  DIVISOR. 

ART.  93.  A  common  divisor  of  two  or  more  numbers,  is 
a  number  that  will  divide  each  of  them  without  a  remain- 
der; 3  is  a  common  divisor  of  12  and  18. 

The  greatest  common  divisor  of  two  or  more  numbers,  is 
the  greatest  number  that  will  divide  each  without  a  re- 
mainder; 6  is  the  greatest  common  divisor  of  12  and  18. 

REMARK. — 1.  Two  numbers  may  have  several  common  divis-.ri, 
ut  tlu-y  cnu  have  only  one  grratest  common  divisor. 
2.  The  greatest  common  divisor  is  often  termed  the  greatett  corv 
mon  measure,  or  greatest  common  factor. 

REVIKW. — 92.  What  is  the  rule  for  finding  all  tho  divisors  of  a 
number?  Why  can  there  bo  no  other  divisor  tbnn  these?  93.  What  ii 
a  common  divisor  of  several  numbers?  tho  greatest  common  divisor/ 
Qivo  examples. 


GREATEST   COMMON   DIVISOR.  01 

ART.  94.  To  find  tho  greatest  common  divisor  of  twc 
numbers, 

HULK  I. — Resolve  the  given  numbert  into  their  prime  factort; 
the  product  of  the  factors  common  to  both  numbers  will  be  the 
greatest  common  divisor  sought. 

Find  the  greatest  common  divisor  of  30  and  105. 

30  =  2X3X5.)    3  X  5  =  15,  tho  greatest 
105  =  3x5x7.  }          common  divisor. 

E  EMONSTK  AT  ION. — The  product  3  X  6  is  a  divisor  of  both  the 
cumbers,  since  each  contains  it;  and  it  is  their  greatest  common 
divisor,  since  it  contains  all  the  factors  common  to  both. 

FIND   THE   GREATEST   COMMON   DIVISOR 

1.  Of  30  and  42 Ans.  2x3=    6. 

2.  Of  42  and  70 Ans.  2  X  7  =  14. 

3.  Of  63  and  105      ....      Am.  3  X  7  =  21. 

4.  Of  66  and  165      ....    Ans.  3  X  11  =  33. 

5.  Of  90  and  150      ...   Ans.  2  X  3  X  5  =  30. 

The  greatest  common  divisor  contains,  as  factors,  all  the  other 
common  divisors;  thus,  30,  which  is  2X3X5,  contains  2,  3,  5, 
2  X  3  =  6,  2  X  5  =  10,  and  3  X  6  =  15,  the  only  remaining  common 
divisors  of  90  and  150. 

6.  Of  60  and  84  .  .  .  Ans.  2  X  2  X  3  =  12. 

7.  Of  90  and  225  .  .  .  Ans.  3  X  3  X  5  =45. 

8.  Of  112  and  140.  .  .  Ans.  2  X  2  X  7  =  28. 

The  greatest  common  divisor  of  more  than  two  numbers  may  be 
found,  by  resolving  each  into  its  prime  factors,  and  taking  the 
product  of  the  factors  common  to  all. 

9.  Of  30,     45,  and    75     .     .      Ans.  3x5  =  15. 
10.     Of  84,  126,  and  210     .   Ans.  2  X  3  X  7  =  42. 

ART.  95.  Rule  1  is  generally  used  \vhcn  the  number* 
are  small ;  but  \vhen  they  are  large,  apply 

RULE  II. — Divide  the  greater  number  by  the  less,  and  the  divisor 
by  the  remainder,  and  so  on;  always  dividing  the  last  divisor  by 
the  last  remainder,  till  nothing  remains;  the  last  divisor  wi'.l  6« 
the  greatest  common  divisor  sought. 


RAY'S   HIGHER   ARITHMETIC. 


Find  the  greatest  common  diviaor  of  24  and  66 

S  o  L  u  T  i  o  N.  —  Divide  CG  by  24  ;  the  '    24)66(2 
quotient  is  2,  and    the   remainder  18.  ^Q 

Next,  divide  24  by  18;  the  quotient  is  —  - 

1,  and  the  remainder  6.     Lastly,  divide  '7o' 

18  by  0;  there  is  no  remainder;  hence, 

€  ia  the  greatest  common  divisor  of  24  6  )  1  8  f  3 

o  1  66.  18 

Rule  2  depends  on  the  following 
PRINCIPLES. 

1.  A  divisor  of  a    number  is   a   divisor  of  any  multiple 
of  that  number.     As  shown  in  Art.  89,  Principle  1st. 

2.  A  common  divisor  of  two  numbers  is  a  divisor  of  their 
SUM.     As  shown,  Art.  89,  Principle  2d. 

3.  A  common  divisor  of  two  numbers  is  a  divisor  of  their 
DIFFERENCE. 

DEMONSTUATION.  —  Since  each  of  the  numbers  contains  the  com- 
mon divisor  a  certain  number  of  times,  their  difference  must  contain 
it  as  many  times  as  the  larger  contains  it  more  times  than  the 
smaller;  2  being  a  divisor  of  16  and  10,  must  be  a  divisor  of  their 
difference;  for,  16  is  8  twos,  and  10  is  5  twos,  and  their  difference  is 
8  twos  minus  5  twos  =  3  twos. 

4.  Tlie  greatest  common  divisor  of  two  numbers  is  also  the 
greatest  common  divisor  of  the  smaller,  and  their  remainder 
after  division. 

OBSERVE,  that  in  the  following  demonstration,  G.  C. 
D.  signifies  greatest  common  divisor. 

DEMONSTRATION.—  The  G.  C.  D.  of  24     24)66(2 
and  G6  divides  24,  and  must,  therefore,  4  8 

divide  48,  which  is  2  times  24,  (Princi-  -,  Q.  ^  .  ,-. 

pie  1st);  as  it  divides  both  66  and  48,  IS 

it  must  divide  their  difference  18,  (Prin-  _ 

ciple  3i);   therefore,    the  G.   C.   D.   of  6)18(3 

24    and  66,  is  a  common  divisor  of  24  18 

od  18. 

It  now  remains  to  show  that  it  is  their  greatest  common  divisor. 

If  there  could  be  any  greater  common  divisor  of  24  and  18,  aa  ii 
would  divide  24,  it  would  also  divide  48  (Principle  1st);  and  as  it 


REVIEW. — 94.  What  is  Rule  1  for  finding  the  greatest  common  divi.«oi 
of  two  numbers  ?  Prove  it.  What  are  contained  in  the  greatest  common 
divisor  ?  How  is  the  rule  applied  to  more  than  two  number*  T 


GREATEST   COMMON   DIVISOR.  03 

jvould  divide  both  48  and  18,  it  would  divide  their  sum  66,  (Principle 
2<1),  and  would,  consequently,  be  a  common  divisor  of  24  and  66.  We 
should  then  have  a  common  divisor  of  24  and  66  greater  than  their 
greatest  common  divisor,  which  is  abeurd.  Hence,  the  G.  C.  D.  of 
24  and  66  is  not  only  a  com.  divisor  of  24  and  18,  but  is  their  0.  C. 
D.;  and  the  proposition  is  proved. 

DEMONSTRATION   OF  RULE   II. 

The  reason  of  the  rule  follows  immediately  from  Principle  4th; 
for,  as  the  0.  C.  D.  of  the  two  given  number's  is  the  same  as  the  Q. 
C.  D.  of  the  smaller  and  their  remainder  after  division,  and  as  this 
is  the  same  as  the  G.  C.  D.  of  the  smaller  of  those  two  and  their  re- 
mainder after  division,  and  so  on ;  it  follows,  that  when  we  get  the 
G.  C.  D.  of  any  remainder  and  the  previous  divisor  (which  is  always 
the  smaller  of  the  two  numbers  used  in  the  division),  this  will  also 
be  the  G.  C.  D.  of  the  original  two  numbers ;  but,  whenever  a  re- 
mainder is  exactly  contained  in  the  previous  divisor,  it  will  neces- 
sarily be  the  G.  C.  D.  of  those  two,  since  no  number  can  have  a  greater 
divisor  than  itself;  and,  therefore,  whenever  this  exact  division 
takes  place,  the  divisor  will  be  the  G.  C.  D.  not  only  of  the  two  num- 
bers last  used,  but  also  of  the  two  numbers  first  given. 

NOTES. — 1.  To  find  the  G.  C.  D.  of  more  than  two  numbers,  first 
find  the  G.  C.  D.  of  any  two;  then  of  that  G.  C.  D.  and  one  of  the 
remaining  numbers,  and  so  on  for  all  the  numbers;  the  last  C.  D. 
will  be  the  G.  C.  D.  of  all  the  numbers. 

2.  If  two  given  numbers  be  divided  by  their  G.  C.  D.  the 
quotients  will  be  prime  to  each  other. 

FIND   THE   GREATEST   COMMON   DIVISOR  OF 


1. 

ANS. 

85  and  120  .  .  5. 

ANS. 

7.  597  and  897  .  .  .  3. 

2. 

91  and  133  .  .  7. 

8.  825  and  1287  .  .  33. 

3. 

357  and  525  .  .  21. 

9.  423  and  2313  .  .  9. 

4. 

425  and  493  .  .  17. 

10.  18607  and  24587  .  23. 

5. 

324  and  1161.  .27. 

11.  105,  231  and  1001.  7. 

G. 

589  and  899  .  .  31. 

12.  165,  231  and  385  .  11. 

13. 

816,  1360,  2040  and  4080   ....  AM.  136. 

14. 

1274,  2002,  2366,  7007  and  13013  .  Ant.  91. 

REVIKW.— 95.  When  is  Rule  1  generally  used?  What  is  Rule  2? 
Ilhiftrato.  What  are  the  1st  and  2d  principle*  on  which  the  demonstration 
of  the  rule  depends  ?  The  3d  ?  Prove  it.  The  4th  ?  Prove  it.  Demon- 
strate »io  rule 


04  RAY'S  HIGHER  ARITHMETIC. 

LEAST  COMMON   MULTIPLE. 

ART.  98.  A  common  multiple  of  two  or  more  numbers, 
is  a  number  tbat  can  be  divided  by  each  of  them  without  a 
remainder;  24  is  a  common  multiple  of  3  and  4. 

The  least  common  multiple  of  two  or  more  numbers,  is 
the  least  number  that  is  divisible  by  each  of  them.  12  is 
the  least  common  multiple  of  3  and  4. 

II  EM  ARK  s. — 1.  Since  the  common  multiple  of  two  or  more  num- 
bers contains  each  of  them  as  a  factor,  every  common  multiple  it  m 
composite  number. 

2.  Since  the  continued  product  of  two  or  more  numbers  is  divisi- 
ble by  each  of  them,  a  common  multiple  of  two  or  more  numbers 
may  always  be  found  by  taking  their  continued  product;  and  since 
any  multiple  of  this  product  will  be  divisible  by  each  of  the  num- 
bers, (Art.  89,  Principle  1st),  an  unlimited  number  of  common 
multiples  may  be  found  for  the  same  numbers. 

TO   FIND   THE   LEAST  COMMON   MULTIPLE. 

ART.  97.  RULE  I. — Separate  the  given  numbers  into  theur 
prime  factors;  then  multiply  together  such,  and  only  such,  ofthote 
factors,  as  are  necessary  to  form  a  product  that  will  contain  all 
the  prime  factors  of  each  number. 

Find  the  least  common  multiple  of  10,  12,  15. 

SOLETION. — Factor  the  num-          10  =  2X5 

bers;  the  prime  factor  2  occurs          -.  ^ O  y  9  y  5 

once  in  10 and  twice  in  12;  strike          -  ^        ~       ~t 
out  the  first  2  and  reserve  2X2 

as  factors  of  the  least  common     2x2x3x5  =  60.  Ant. 
multiple.     The  factor  5   occurs 

once  in  10  and  once  in  15;  strike  out  the  first  6,  and  reserve  the 
second  as  a  factor  of  the  least  common  multiple.  The  factor  3  occurs 
once  in  12  and  once  in  15;  strike  out  the  first  3,  and  reserve  the 
second  3  as  a  factor  of  the  least  common  multiple.  Lastly,  take  the 
product  of  the  reserved  prime  factors  2X2X3X5  =  00,  for  the 
cast  common  multiple. 

DEMONSTRATION. — The  product  2X2X3X5  =  60,  is  a  corn- 
Dion  multiple,  because  it  contains  all  the  factors  of  each  of  the 

REVIEW. — 96.  What  is  a  common  multiple  of  serernl  numbers?  What 
is  the  least  common  multiple?  Give  examples.  How  many  common  mul- 
tiples may  ho  found  for  several  numbers?  Why?  97.  Give  Rule  1.  Prove 
it.  How  often  must  a  factor  be  taken  in  the  least  common  multiple?  Why? 


LEAST  COMMON  MULTIPLE.  65 

numbers;  it  is  the  least  common  multiple,  because  it  does  not  contain 
any  prime  factor  not  found  in  some  one  of  the  numbers. 

REMARKS. — Each  factor  must  be  taken  in  the  least  common  mul- 
tiple the  greatest  number  of  times  it  occurs  in  either  of  the  numbers. 
In  the  preceding  solution,  2  must  be  taken  twice,  because  it  occurs 
twice  in  1'2,  the  number  containing  it  most. 

2.  To  avoid  mistakes,  after  resolving  the  numbers  into  their  prime 
factors,  strike  out  the  needless  factors. 

FIND   TUB   LEAST  COMMON    MULTIPLE   OP 


2.  8,  10,  15    .  120. 

3.  G,    9,12    .     30. 

4.  12,  18,  2-4    .    72. 


5.  8,14,21,   28    .   168. 

6.  10,15,20,    30    .     CO. 

7.  15,30,70,105    .  210. 


ART.  98.  RULE  II.  —  1.  Place  the  numbers  in  a  tine;  strike 
out  any  that  will  exactly  divide  any  of  the  others;  divide  by  any 
prime  number  that  will  divide  two  or  more  of  them  without  a 
remainder;  set  the  quotients  and  undivided  numbers  in  a  line 
beneath. 

2.  Proceed  with  this  line  as  before,  and  continue  the  operation 
till  no  number  greater  than  1  will  exactly  divide  two  or  more  of 
the  numbers. 

3.  Multiply  together  the  divisors  and  the  numbers  in  the  last 
line;  their  product  will  be  the  least  common  multiple  required. 

Find  the  least  common  multiple  of,  10,  20,  25  and  30. 

SOLUTION.  -Strike       2)J0         2  Q         ^         30 

out'  the  10,  because  it  is          '  ^r 


exactly  contained  in  20;       5)  10,        25,        15 

then   divide   by   2,   be-     -  -  ^  -  £  —     —  g 

cause  it  is  a  factor  of 

two  of  the  numbers,  20       2X5X2X5X3  =  300  Am. 

and  30.    Next  divide  by 

6,  because  it  is  a  factor  of  more  than  one  of  the  remaining  numbers. 

Lastly,  multiply  together  the  divisors  2  and  5,  and  the  remaining 

numbers,  2,  6  and  3;  their  product,  300,  is  the  least  com.  multiple. 

DEMONSTRATION.  —  The  10  is  marked  out  to  shorten  the  opera- 
tion. The  least  common  multiple  of  20,  25  and  30,  contains  20,  and 
therefore  contains  10,  which  is  a  factor  of  20,  and  will  be  th#  least 

RKVIKW.—  98.  Give  Rule  2.     Prove  it.     What  kind  of  numbers  muM 
lie  used  as  divisors  ?    Why  ?    When  the  numbers  are  prim»  to  each  othe* 
how  ig  the  least  common  multiple  found  ? 
6 


60  RAY'S   HIGHER   ARITHMETIC. 


common  multiple  of  all  the  numbers.  The  rest  of  the  demonstration 
is  similar  to  the  previous  one;  for  the  division  by  the  prime  numbers 
serves  to  strike  out  the  needless  factors,  and  those  divisors  with  the 
factors  remaining  in  the  last  line,  are  evidently  the  reserved  fac- 
tors necessary  to  constitute  the  least  common  multiple.  Fora  full 
analysis,  see  Ray's  Arithmetic,  Third  Book,  Art.  120. 

NOTE. — In  dividing  to  cancel  the  needless  factors,  divide  by  a 
frime  number.  Dividing  by  a  composite  number  would  not,  in  all 
eases,  cancel  all  the  needless  factors;  in  the  preceding  example,  if 
we  divide  first  by  10,  the  5  in  the  number  25  would  not  be  canceled. 

ART.  99.  RULE  III. — 1.  Set  the  numbers  in  a  line,  striking 
out  such  as  are  contained  in  any  of  the  others,  and  separate 
any  convenient  one,  generally  the  largest,  from  the  others,  by  a 
curved  line. 

2.  Divide  each  of  the  remaining  numbers  by  the  greatest  divisor 
common  to  it  and  the  number  cut  off,  and  set  the  quotients  in  a 
line  beneath. 

3.  Proceed  with  the  second  line  exactly  as  with  the  first,  and 
continue  so  until  all  the  quotients  are  observed  to  be  prime  to  each 
other;  the  continued  product  of  these  quotients  and  the  numbers 
cut  off",  will  be  the  least  common  multiple  required. 

NOTE  8. — 1.  If  the  number  cut  off  is  found  to  be  prime  to  all  the 
rest  in  the  same  line,  cut  off  another,  and  proceed  with  it,  reserving 
the  first  as  a  factor  of  the  least  common  multiple. 

2.  When   the  given  numbers  contain  no   common   factor,  it    ia 
evident  their  product  will  be  the  least  common  multiple  required ; 
the  least  common  multiple  of 

4,  9  and  25,  is  4  X  9  X  25  =  900. 

3.  The  least  common  multiple  of  several  numbers  is  equal  to  the 
product  of  their  greatest  common  divisor,  by  those  factors  of  each 
number  not  found  in  the  others. 

4.  If  the  least  common  multiple  of  several  numbers  be  divided  by 
cither  of  them,  the  quotient  will  be  the  product  of  all  the  factors  of 
the  others  not  found  in  the  divisor. 

Resume  the  example  under  Rule  2. 

SOLUTION.— After  striking  out  OPERATION.^ 

the  10,  cut  off  30  from  the  rest  by       %  0,  20,       25(30 

a  curve,  and  then  divide  20  by  10,  5 

which  is  the  greatest  divisor  com-      ~        _       „« o  n  A 

mon  to  it  and  30 ;  also  divide  25  by      ^  X  D  X  3  U  - 

6,  the  greatest  divisor  common  to  it  and  30.  The  quotients  are  set 


CASTING  OUT   THE   NINES.  07 


beneath,  and  being  prime  to  each  other,  further  cutting  off  and  di- 
viding are  unnecessary,  and  the  product  of  2,  5  and  30,  gives  300, 
the  least  common  multiple  required. 

DEMONSTRATION. — The  least  common  multiple  must  contain  th« 
number  cut  off  (30),  and  such  factors  of  the  rest  as  are  not  found  in 
the  30;  011  this  account,  divide  each  of  them  by  the  greatest  factor 
common  to  it  and  the  ntlmber  cut  off,  thus  getting  rid  of  the  neediest 
factors;  in  like  manner,  the  least  common  multiple  must  contain 
the  number  cut  off  in  the  second  line,  and  such  factors  of  the  rest  at 
are  not  found  in  the  one  cut  off;  therefore,  we  repeat  the  process 
for  each  line,  until  all  the  numbers  in  the  same  line  are  found  to  be 
prime  to  each  other;  when  this  is  the  case,  there  will  be  no  more 
needless  factors,  and  the  least  common  multiple  will  be  the  continued 
product  of  the  numbers  cut  off  and  those  in  the  last  line. 

FIND   THE   LEAST   COMMON   MULTIPLE  OP 

ANS.  I  A.N'S. 


1.  4,6,15  ...    60. 

2.  6,9,20   .    .    .  180. 


3.  15,20,30   .    .    60. 

4.  7,11,13,5  .    5005. 


5.  35, 45, 63,  TO     .630. 

6.  10,14,20,35     .140. 

7.  8,15,20,25,30.600. 

8.  15,24,40,140  .  840. 


9.  30,45,48,80,120,135     ....    AM.  2160. 

10.  77,91,105,143,165,195     .    .    .  4iw.  15015. 

11.  174,485,4611,14065,15423  .  Am.  4472670. 

12.  498, 85988, 235803, 490546.  AM.  244291908. 


PROOF   OF    THE  ELEMENTARY  RULES 

BY   CASTING    OUT   THE   NINES. 

ART.  100.  To  cast  the  9's  out  of  any  number,  is  to 
divide  the  sum  of  its  digits  by  9,  and  find  the  excess.  To 
do  this,  begin  at  the  left,  add  the  digits  together,  and 
when  the  sum  is  nine  or  more,  drop  the  9,  and  carry  the 
excess  to  the  next  digit,  and  so  on. 

For  example,  to  cast  the  9's  out  of  768945,  say  7  and 
6  arc  13,  which  is  4  above  9;  drop  the  9,  and  carry  the  4; 

REVIEW. — 99.  Give  Rule  3.  If  a  number  cut  off  is  primo  to  the  rest 
in  tho  same  lino,  what  must  be  done  ?  Prove  the  rule.  100.  What  it 
meant  by  casting  the  9's  out  of  any  number  ? 


68  RAY'S    HIGHER   ARITHMETIC. 


4  and  8  arc  12,  which  is  3  above  9;  then,  3  (the  digit  9 
beiri£  passed  over)  and  4  are  7  and  5  arc  12,  which  13  3 
above  9  ;  consequently,  the  excess  in  this  instance,  ia  3 

All  the  methods  of  proof  arc  founded  on  this 

PRINCIPLE. — Any  number  divided  by  9  will  leave  the 
tame  remainder  as  the  sum  of  its  dijit^  divided  l>ij  9. 

For  example,  take  345G. 

f  3000=3(1000)=3x(999+l)=3x999+3 

q  400=  4(100)=  4x(99-f-l)=  4x99+4 

50=     5(10)=     5X(9+1)=     5x9+5 

[       G= 0 

DEMONSTRATION. — Observe  that  3000  is  a  certain  number  of 
9's,  with  a  remainder  3;  400,  a  certain  number  of  9's,  with  a  re- 
mainder 4 ;  50,  a  certain  number  of  9's,  with  a  remainder  5;  and  C, 
being  less  than  9,  may  be  termed  a  remainder.  The  remainders, 
3,  4,  5,  6,  are  the  digits  of  the  number  3450;  hence,  the  excess  of  9'a 
in  the  number  3456  is  the  same  as  that  in  its  digits,  3,  4,  5  and  C. 

PROOF   OF   ADDITION. 

Cast  the  l)'s  (or  ll's)*  out  of  each  of  the  numbers  added,  also 
out  of  their  sum ;  the  last  excess  must  equal  the  sum  of  the  other* 
after  dropping  all  9's  (or  ll's). 

*  To  cast  the  ll's  out  of  any  number,  see  page  70. 

The  excesses  in  the  numbers  are  8,  2,  4  ^  Q  „  _        EXCESS. 

and  3,  and  the  excess  in  the  sum  of  these  r.QO~i 

excesses  is  8.     The  excess  in  the  sum  of 
the  numbers  is  8,  the  two  excesses  being  J?  OAO 

the  same,  as  they  ought  to  be  when   the  i  o  U  ^ 

work  is  correct.  20729  8 

NOTE. — In  proving  Addition  by  this  method,  it  is  not  necessary 
to  stop  at  each  number  and  write  the  excess,  but  regard  all  the 
numbers  as  forming  one  horizontal  line. 

DEMONSTRATION. — Divide  each  number  by  9,  and  add  up  the 
remainders,  omitting  the  9's,  if  any;  the  result  must  be  the  same  aa 
the  remainder  obtained  by  dividing  the  sum  of  the  numbers  by  9; 
and  as  these  remainders  are  most  easily  obtained  by  cnsiing  the 
9's  out  of  each  number,  (see  Principle),  the  reason  of  the  proof  is 
apparent. 

REVIEW. — 100.  On  what  principle  do  the  proofs  by  ousting  out  the  9'i 
depend  ?  Prove  it.  What  is  the  proof  of  addition  ? 


CASTING   OUT  ELEVENS.  69 

PROOF   OF    SUBTRACTION. 

Cast  the  9's  (or  11'*)  out  of  the  subtrahend,  the  remainder, 
and  the  minuend;  the  last  excess  will  be  equal  to  the  sum  of  the 
other  two,  after  dropping  all  9'a  (or  ll's). 

EXAMPLE. — Proceeding  ns  directed  in  h 

de  note  to  last  proof,  we  find  the  excess  Minuend,       76 

In  the  subtrahend  and  remainder  together  Subtrahend,  1 

to  be  8:  the  excess  in  the  minuend  is  also  Remainder     G  4  0  6 
8,  as  it  should  be  when  the  work  is  right. 

DEMONSTRATION. — As  the  minuend  is  the  sum  of  the  subtra- 
hend and  remainder,  the  reason  of  this  proof  is  seen  from  that  of 
Addition. 

PROOF   OF   MULTIPLICATION. 

Cast  the  9's  (or  ll's)  out  of  the  multiplicand  and  multiplier; 
multiply  the  two  excesses  together;  cast  (he  9's  (or  1 1 'a)  out  of  the 
result,  and  also  out  of  the  product;  when  the  work  is  correct,  the 
last  two  excesses  will  agree. 

Multiply  835  by  70;  the  pro-  8  q  ^  _  QOyO-4-7 

duct  is  G34GO.     The  excess  in  °  2  Ji  oCn" 

the  multiplicand  is  7,  in  the 

multiplier  4,  and  in  the  pro-  3G8  X  0  -f-  2  8 

duct  1;  the  two  former  multi-  730  X  81   4-   5  G_X_9 

plied  give  28,  and  the  excess  ^SlTx^l  4-424  X~9  4-28 
\n  28  is  also  1,  as  it  should  be. 

DEMONSTRATION. — Since  each  of  the  numbers  to  be  multiplied 
contains  a  certain  number  of  9's  and  an  excess,  their  product,  as 
seen  from  the  work,  will  consist  of  three  parts,  two  of  which  are 
divisible  by  9;  therefore,  the  excess  of  9's  in  the  product  must  be 
the  same  as  the  excess  of  9's  in  the  3d  part,  (IS),  which  is  the 
product  of  the  excesses  in  the  multiplicand  and  multiplier;  nnd 
siniie  these  excesses  are  most  easily  obtained  by  casting  the  9's  out 
of  the  digits  of  each  number,  (see  Principle),  the  reason  of  the  proof 
is  apparent. 

PROOF  OF  DIVISION. 

Cast  the  9's  (or  1 1's)  out  of  the  divisor,  dividend,  quotient,  and 
rimainder.  Multiply  together  the  excesses  in  the  dicisor  and  quo- 
tient, and  cast  the  9's  (or  ll's)  out  of  the  result;  then,  to  this 
excess,  add  the  excess  in  the  remainder,  and  cast  the  9'*  (or  }  1  '*) 
911 1  of  (he  result ;  when  the  work  is  correct,  this  excess  will  be  tht 
tame  as  the  excess  in  the  dividend. 

REVIEW. — What  is  the  proof  of  subtraction?     Of  multiplication T 


70  RAY'S  HIGHER  ARITHMETIC. 

jMv;de  8015  by  25;  the  quotient  is  356  and  the  re- 
m.'iLidiT  15. 

Kyi  ess  of  9's  in  the  divisor     .......  7. 

Ex  vss  of  9's  in  the  quotient 5. 

1  X  5  =  35.     Excess  of  9's  in  35 & 

jRxcesis  of  9's  in  the  remainder 6. 

6  -L  8  =  14.     Excess  of  9's  in  14 5^ 

I?xcoss  of  9's  in  the  dividend  is  also      ....  5. 

DEJIONSTAATION. — Since  the  dividend  is  the  product  of  the 
divi'orund  quotient,  with  the  remainder  added,  the  reason  of  this 
prooi  <s  seen  from  those  of  Multiplication  and  Addition. 

-A 

PROOF  BY  CASTING  OUT  THE   ELEVENS. 

'To  cast  the  ll's  out  of  any  number,  add  its  alternate 
film's,  commencing  at  the  right,  and  dropping  11  when 
ths:  sura  exceeds  that  number;  then  do  the  same  with  the 
remaining  figures,  and  subtract  the  last  result  from  the 
former,  increased  by  11  if  necessary. 

For  example,  to  cast  the  ll's  out  of  30752486,  say  6 
and  4  are  10  and  5  are  15;  drop  11  and  4  remains, 
th^n  4-  and  0  are  4;  also,  8  and  2  are  10  and  7  are  17, 
droj  1J  and  6  remains,  and  6  and  3  are  9;  9  from  4  can 
not  bt>  taken,  but  4  increased  by  11  is  15,  and  9  from  15 
leaves  6,  which  is  the  excess  required. 

Tha  Proofs  by  casting  out  the  ll's  depend  on  this 
PRINCIPLE. — Any  number  divided  by  11  leaves  the  same 
remct'ixder  as  the  excess  obtained  by  casting  out  its  11'*. 

Ti'8  principle,  and  the  proofs  to  which  it  leads,  can  be 
dexrou^trated  in  a  manner  similar  to  those  regarding  the 
9's,  t?nd,  by  putting  11  for  9  in  the  proofs  by  9's,  we 
ba\v  tbe  proofs  by  ll's.  (See  Proofs.) 

RESL  \HK. — These  methods  of  proof  are  liable  to  this  objection; 
two  Sguies  may  be  substituted  for  each  other,  or  the  correct  figurei 
may  ':/e  replaced  by  wrong  ones  having  the  same  sum,  and  the  work 
appear  to  prove  when  it  is  wrong.  These,  however,  occur  so  rarely 
aa  nov  to  detract  much  from  the  merits  of  the  methods. 

R.  v  *"1K  w. — What  is  the  proof  of  division?  Illustrate  and  give  the 
reason  XT  these  proofs.  100.  Explain  the  proofs  by  casting  out  the  ll'i. 
Whei  will  these  proofs  fail  to  detect  an  error? 


CANCELLATION.  71 


CANCELLATION. 

ART.  101.  CANCELLATION  is  a  short  method  of  obtain- 
ing results  by  omitting  equal  factors  from  a  dividend  and 
divisor.  Its  use  is  seen  in  the  following  examples. 

Exchanged  24  hats  at  $5  each,  for  coats  at  $24  each: 
how  many  coats  did  I  get? 

SOLUTION.  —  To  solve  this  question,  multiply  24  by  5,  and  divide 
the  product  by  24.  Instead  of  performing  this  work,  indicate  it 

24X  5 

thus,  -         —  ;   then,  since  dividing  either  factor  of   a  product 

it 

divides  the  product,  (Art.  71),  the  result  is  1  X  6  =  5  ;  the  same  aa 
would  be  got  by  canceling  the  24  from  both  dividend  and  divisor. 

Multiply  105  by  18,  and  divide  the  product  by  30. 

SOLUTION.  —  Indicate  the  operations  as 

in  the  margin.      Divide  both  dividend  and     2  1  j  ,,  -  ~.  *  o  3 
divisor  by  6;   this  gives  105  X  3  (Art  71,)  AVP  *  f-P 

above,  and  5  below,  and  docs  not  alter  the  $  P  K 

quotient,  (Art.  75).     The  quotient  is  found,       Q-.         o  _  pq 
as  in  last  example,  to  be  21  X  3  (Art.  71)  or       ~      X 
63.     As  each  factor  is  used  in  canceling,  it  is  crossed  to  indicate 
that  there  is  no  further  use  of  it;  and  each  quotient  is  placed  beside 
the  number  from  which  it  is  obtained.     The  21  and  3  being  the  fac- 
tors left  of  the  dividend  after  cancellation,  are  multiplied  together; 
their  product  is  03,  and  as  no  factor  of  the  divisor  is  left,  the  63  is 
not  to  be  divided,  and  is,  therefore,  the  quotient  required. 

Multiply  75,  153  and  28  together,  and  divide  by  the 
product  of  63  and  36. 

SOLUTION.  —  Indicate   the  25        17  * 

operations  as  in   the  margin.  JT$X^£$X$!$^ 

Cancel  4  out  of  28  and  36,  leav-  3  X  3  fi 


ing  7  above,  and  9  below.  Can- 
eel  this  7  out  of  the  dividend 


Q  3 

and  out  of  the  63  in  the  di-       2  5  X  1  7  _  4  2  5  _  ..  .  - 

visor,  leaving  9  below.    Cancel  o 

i  9  out  of  the  divisor,  and  out 

of  153  in  the  dividend,  leaving  17  above.     Cancel  3  out  of  9  and  75, 

leaving  25  above  and  3  below.     No  further  canceling  is  possible; 

the  factors  remaining  in  the  dividend  are  25  and  17,  whose  product. 

425  divided  by  the  3  in  the  divisor,  gives 


72  RAVS  HIGHER  ARITHMETIC. 

Anr.  102.  If  the  dividend  or  divisor,  or  both,  is  the 
product  of  several  factors,  the  quotient  can  often  be 
easily  obtained  by  this  ^ 

RULE   FOR   CANCELLATION. 

1.  Indicate  the   mid  I  i plications    which  produce    the    dividend, 
and  those,  if  any,  which  produce  the  divisor. 

2.  Cancel  equal  factors  from  dividend  and  dicisor;    multiply 
together   the  factors  remaining   in   the  dividend,  and   divide  the 
product  by  the  product  of  the  factors  left  in  the  divisor. 

NOTES. — 1.  If  no  factor  remains  in  the  divisor,  the  product  of 
the  factors  remaining  in  the  dividend  will  be  the  quotient;  if  only 
one  factor  is  left  in  the  dividend,  it  will  be  the  answer. 

2.  Cancellation  can  only  take  place  between  a  factor  of  the  divi- 
dend and  a  factor  of  the  divisor;  not  between  two  factors  of  the 
dividend,  or  two  factors  of  the  divisor. 

3.  Canceling  equal  factors  is  dividing  both  dividend  and  divisor 
by  the  same  number,  which  does  not  atl'ect  the  value  of  the  quotient, 
(Art.  75.) 

EXAMPLES   FOR  PRACTICE. 

1.  How  many  cows,  worth.  $24  each,  can  I  get  for  9 
horses  worth  §&0  each?  Ans.  30. 

2.  I  exchanged  8  barrels  of  molasses,  each  containing 
33   gallons,  at  40   cents  a  gallon,   for   10   chests   of   tea, 
sach  containing  24  pounds:    how  much  a  pound  did  the 
tea  cost  me?  Ans.  44  cents. 

3.  How  many  bales  of  cotton,  of  400  pounds  each,  at 
12   cents    a    pound,    arc    equivalent    to    G    hogsheads   of 
mgar,  000  pounds  each,  at  8  cents  a  pound?       Ans.  9. 

4.  Divide  15  X  24  X  112  X  40  X  10  by  25  X  30  X  5G 
X  90.  Ans.  32 


VIII.   COMMON  FRACTIONS. 

- 

ART.  103.  A  fraction  is  a  part  of  a  unit  or  whole  thine, 
when  it  is  divided  into  rqnal  parts.  If  an  apple  is  divided 
into  3  equal  parts,  each  part  is  a  fraction  of  the  apple. 

NOTE.  —  Fraction  is  derived  from  the  Lniin/ractus,  broken. 

REVIEW. — 101.  What  is  Cancellation?  Illustrate  it.  102.  In  what 
oase  may  cancellation  bo  employed?  Give  tho  rule. 


COMMON   FRACTIONS.  73 


ART.  104.  The  name  of  d  fraction^  ami  tJir  size  nf  itt 
p<7/7x,  ifi'/ii-ni/  nn  fit/1  tn/ntbrr  of  f,arfn  !nln  «•  A  /'••/(  <hf  unit 
t'x  ilivii/i-il  ;  thus,  whon  fhc  unit  is  divided  :*ito  '2,  3,  or  4 
Ci|ii:il  parts,  the  fractions  arc  named  Imfce*,  Hiinl*  and 
f'<ur'h*  (Art.  <J2):  moreover,  a  /<«///'  is  evidently  larger 
than  a  third,  a  third  than  &  fourth,  and  soon. 

A«r.  105.  Fractions  are  divided  into  two  classes. 
ibntmoK  and  Decimal. 

A.  common  fraction  is  expressed  by  two  numbers,  one 
above  the  other,  with  a  horizontal  line  between  them;  onc- 
1)  ill'  is  expressed  by  A,  two-thirds  by  f. 

The  number  below  the  line  is  called  the  denominator, 
kfir-.-ui.se  it  denominates  or  gives  name  to  the  fraction:  it 
shows  into  how  many  parts  the  unit  is  divided. 

The  number  above  the  line  is  called  the  numerator,  be- 
cause it  a  inn  hers  the  parts  :  it  shows  how  many  parts  the 
fraction  contains  ;  in  the  fraction  fj  the  denominator  7, 
shows  that  the  milt  contains  seven  equal  parts;  the  nume- 
rator 4,  shows  that  the  fraction  contains  4  of  those  parts. 

The  terms  of  a  fraction  are  the  numerator  and  denomi- 
nator; the  terms  of  1  are  5  and  8. 

ART.  106.  There  are  two  methods  of  considering  a 
fraction  whose  numerator  is  greater  than  1;  thus,  to  divide 
2  apples  of  the  same  size,  equally  among  3  boys,  divide 
each  apple  into  three  equal  parts,  making  6  parts  in  all, 
and  then  give  to  each  boy  two  of  the  parts,  expressed  by  |. 
The  2  parts  may  be  taken  from  one  apple,  or  1  part  from 
each  of  the  fico  apples;  hence,  5  expresses  cither  2  thirds 
of  one  thing,  or  1  third  of  two  things.  Also,  5  expresses 
cither  4  fifths  of  one  thing  or  1  fifth  of  four  things  ; 
therefore, 

The  numerator  of  a  fraction  may  le  regarded  as  showing 
tit.'',  number  of  units  to  Le.  divided  ;  the  denominator,  the  nnm- 
brr  of  parts  Info  which  the  numerator  is  to  be  divided:  the 
fraction,  itxelf  being  the  value  of  one  of  those  parts. 


.  —  102.  If  no  fiv'/or  remains  in  the  divisor,  what  is  the  quo- 
dent?  I'etwien  whiK  factors  only  cnn  CHncellaiion  take  (iliioe?  What 
4oes  canceling  equal  fiictors  amount  to?  lO.'i.  \Vhntisii  Fraction?  Why 
lo  called?  I'M.  How  are  fractions  named?  What  duos  the  value  of  a 
fractional  j>art  depend  on?  105.  How  are  fraction?  divided?  Il«w  nrd 
ftotuinon  friii-iinns  expressed?  What  is  the  denominator?  What  is  tii 
\iumurator?  Why  arc  they  called  so  ?  What  are  both  together  called? 

7 


74  KAY'S   HIGHER   ARITHMETIC. 


Hence,  a  'raction  may  be  considered  as  an  expression  of 
unexecuted  division,  in  which 

The  DIVIDEND  is  called  the  NUMERATOR; 
The  DIVISOR      is  called  the  DENOMINATOR; 
The  QUOTIENT  is  called  the  FRACTION  itself. 

|  is  the  quotient  of  1  (num'tor)  divided  by  4  (denom.) 
|  is  the  quotient  of  3  (num'tor)  divided  by  5  (denom.) 
I  is  the  quotient  of  7  (num'tor)  divided  by  6  (denom.) 

NUMERATION  AND  NOTATION. 

ART.  107.  Since  Fractions  arise  from  Division,  one  of 
the  signs  of  Division,  (Art.  59),  is  used  in  expressing 
them ;  that  is,  the  numerator  is  written  above,  and  the 
denominator  below  a  horizontal  line  ;  hence, 

TO   READ   COMMON   FRACTIONS, 

RULE. — Read  the  number  of  parts  taken  as  expressed  by  the 
numerator,  and  then  the  size  of  the  parts  as  expressed  by  the 
denominator. 

TJ  is  read  seven  ninths. 

REMARK. — Seven  ninths,  (|),  signifies  either  7  ninths  of  one, 
or  ^  of  7,  or  7  divided  by  9,  (Art.  100). 

Fractions  to  be  read;  that  is,  expressed  in  words: 

§,  8,  X,  1,  I?,  J,  14,  _9_,  13,  J_,  11,  8  - 
7  8  10  11  IT  27  41  65  90  100  230  1345 

Of  these  fractions,  which  expresses  parts  of  the  largest 
size?  Which,  the  smallest  size?  Which,  the  least  num- 
ber of  parts?  Which,  the  greatest  number  of  parts? 
Which,  the  same  number  of  parts  ?  Which,  parts  of  the 
same  size  ? 

TO   WRITE   COMMON   FRACTIONS, 

ART.  108.  RULE. —  Write  the  number  of  parts ;  place  a  lion- 
tontal  line  below  it,  under  which  write  the  number  which  indi- 
cates the  size  of  the  parts. 

REVIKW. — 106.  How  nanny  methods  of  considering  a  fraction?  Illus- 
trate them.  What  operation  is  expressed  by  a  fraction  ?  What  i*  tar 
dividend?  the  divisor?  the  quotient? 


COMMON   FRACTIONS.  75 


Fractions  to  be  expressed  in  figures: 

Seven  eighths.  Four  eleventh.  Five  thirteenth*.  Una 
tcvcnfccHth.  Three  twenty-ninths.  Eight  Iwenty-oncth*. 
Nine  forty-tooths.  Nineteen  ninety-thirds.  Thirteen  OM 
hundred-fits.  Twenty-four  one  hundred  and  fifteenths. 

To  express  a  whole  number  in  the  form  of  a  fraction, 
write  1  below  it  for  a  denominator;  thus,  2  =  f,  7  =  f,  &c. 

PROPOSITIONS. 

ART.  109.  PROP.  1. —  When  the  numerator  of  a  fraction 
is  less  than  the  denominator,  the  fraction  is  less  than  1. 

PROP.  2. —  When  the  numerator  of  a  fraction  is  equil  to 
the  denominator,  the  fraction  is  equal  to  1. 

PROP.  3. —  When  the  numerator  of  a  fraction  is  greater 
than  the  denominator,  the  fraction  is  greater  than  1. 

DEMONSTRATION. — Prop.  1  is  true,  because,  in  that  case,  the 
number  of  parts  in  the  fraction  is  less  than  the  number  of  parts 
in  a  unit. 

Prop.  2  is  true,  because,  in  that  case,  the  number  of  parts  in  the 
fraction  is  the  same  as  the  number  of  parts  in  a  unit. 

Prop.  3  is  true,  because,  in  that  case,  the  number  of  parts  in  the 
fraction  is  greater  than  the  number  of  parts  in  a  unit. 

ART.  110.  DEFINITIONS. — 1.  A  proper  fraction  is  one 
whose  numerator  is  less  than  the  denominator;  as,  4j  §• 

2.  An  improper  fraction  is  one  whose  numerator  is  equal 
to,  or  greater  than,  the  denominator;  as,  |  and  g. 

REMARK. —  The  word  fraction  primarily  signifies  a  part.  A 
proper  fraction  is  properly  a  fraction  expressing  a  value  less  (ban 
the  whole.  An  improper  fraction  is  not  properly  a  fraction,  the  value 
expressed  being  equal  to,  or  greater  than,  the  whole. 

3.  A  simple  fraction  is  a  single  fraction,  proper  or  im- 
proper; as,  \,  f.  or  f. 

REVIKW. — 107.  What  is  tho  rulo  for  reading  fractions?  for  writing 
them?  108.  What  two  other  methods  of  reading  a  fraction  aro  there T 
Ilnw  may  9  whole  number  bo  expressed  as  a  fraction?  109.  When  is  a 
fraction  less  than  1?  equal  to  1  ?  greater  than  1?  Give  tho  reasons, 
110.  What  is  a  proper  fraction?  an  improper  fraction?  Why  so 
What  is  a  simple  fraction  T 


•JO  RAY'S    HIGHER   ARITHMETIC. 

4.  A  m, n pound  fraction  is  a  fraction  of  a  fraction,  or 
several  fractious  connected  by  of;  as,  |  of  J  of  i. 

f>.  A  mired  niimlier  is  a  fraction  joined  with  a  whole 
n  u  in  be  r ;  as,  1 .',  and  '2  i . 

0.  A  complex  fraction  is  one  having  a  fraction  cither  in 

•9 '     1        *  1  ! 

one  or  both  of  its  terms;  as,  r-p   -—,  -j- ,  and  '.-J. 

4     ot       5  04 

ART.  111.  To  show  that  «  of  £  =  j ;  that  I  of  A  =  J; 
that  j  of  j  =  T'.j,  arid  so  on. 

DEMONSTRATION. — Divide  a     A . . . B 

line  as  A  B  into  two  equal  parts; 

one  of  the  parts,  as  A  C,  is  termed  one  half  (i):  divide  a  half  into  3 
equal  parts,  us  iu  the  figure;  one  of  the  parts  is  called  one  third  of 
one  half,  which  is  expressed  by  figures  thus,  g  of  .j.  This  ia  evi- 
dently one  sixth  of  the  whole  line,  that  is,  f  of  £••.$.  In  like 
manner,  A  of  3  =  j ;  5  of  3  =  |;  3  of  |  =  y'j,  and  so  on. 

What  is  i  of  i?  Why?  What  is  £  of  \1  {  of  ^? 
What  is  ^  of  |? 

PROPOSITIONS. 

ART.  112.  PROP.  I. — Multiplying  the  numerator  of  a 
fraction  likewise  multiplies  the  fraction.  Multiply  the  nu- 
merator of  the  fraction  5  by  3;  the  result,  |,  is  three  times 
as  great  as  |. 

ART.  113.  PROP.  II.  —  Dividing  (he  numerator  of  a 
fraction  likewise  divides  (he  fraction.  Divide  the  numera- 
tor of  the  fraction  f  by  2;  the  result,  |,  is  only  one  half 
as  great  as  §. 

ART.  114.  PROP.  III. — Multiplying  the  denominator  of 
a  fra rtism,  on  the  contrary,  divides  the  fraction.  Multiply 
the  denominator  of  the  1'raction  ^  by  3;  the  result,  'g*, 
ih  one  third  as  great  as  !2-. 

ART.  115.  PROP.  IV. —  Dividing  the  denominator  of  a 
fraction,  on  the  contrary,  multiplies  the  fraction.  Divide 
the  denominator  of  the  fraction,  'g2,  by  2;  the  result,  *!, 
id  da  ice  as  great  as  'ga. 

REVIEW. — What  i?  a  compound  fraction?  a  mixed  number?  a  com- 
plex fraction?  112.  What  is  the  effect  of  multiplying  the  Dtuuenitor  of 
a  fraction?  113.  Of  dividing  the  numerator?  114.  Of  multiplying  th« 
denominator?  115.  Of  dividing  the  donrminatorT 


COMMON   FRACTIONS.  7-7 


AllT.  116.  PROP.  V. —  Multiplying  loth  term*  of  <i  frac- 
tion In/  /lit  fame,  number,  chaime*  iV.s  form  but  Joes  not  utter 
its  value,.  Multiply  both  forms  of  the  fraction  |  by  3;  the 
result,  5>,  lias  the  same  value  as  |. 

ART.  117.  I*  ROP.  VI.—  Dividing  loth  terms  of  a  frac- 
tion l>y  tltt  same  niinther,  changes  it*  form,  out  Joes  not  alter 
T'JS  {-<il itc.  Divide  both  terms  of  the  i'raction  jj  by  4;  the 
result,  |,  has  the  tame  value  as  |. 

DEMONSTRATION. — Observe  tlmt  Ihe  numerator  of  a  fr-iciion 
is  a  dividend  of  which  (lie  denominator  is  tlie  divisor,  and  the  value 
of  the  fraction  the  quotient.  (Art.  100). 

Prop.  1  is  true,  because,  (Art.  73),  multiplying  the  dividend  by 
any  number,  multiplies  the  quotient  by  the  same  number. 

Prop.  '2  is  true,  because,  (Art.  71),  dividing  the  dividend  by  any 
number,  divides  the  quotient  by  the  same. 

Prop.  3  is  true,  because,  (Art.  7-1),  multiplying  the  divisor  by  any 
nuinlier,  divides  the  quotient  by  the  same. 

Prop.  4  is  true,  because,  (Art.  73),  dividing  the  divisor  by  any 
number,  multiplies  the  quotient  by  the  same. 

Prop.  6  and  6  are  true,  because,  (Art.  75),  multiplying  or  dividing 
both  dividend  and  divisor  by  the  same  number,  the  quotient  remains 
the  same. 

REMARK. — ITence,  there  are  two  ways  of  multiplying  a  fraction, 
two  ways  of  dividing  a  fraction,  and  two  ways  of  changing  it.-  form 
without  altering  its  value;  that  is,  multiplying  or  dividing  the  nu- 
merator tioei  the  tame  to  the  /ruction;  but  multiplying  or  dividmg  the 
denominator  duet  the  oppoiile  to  the  fraction;  tohdt  multiplying  or  di- 
viding both  term*  alike  ma  In  no  change  in  in  value. 


REDUCTION  OF  FRACTIONS. 

ART.  118.  Reduction  of  Fractions  consists  in  changing 
their  forms  without  altering  their  values. 

CASE  I. 
TO   REDUCE   A   FRACTION    TO    ITS   LOWEST   TERMS. 

ART.  119.  A  fraction  is  in  its  lowest  terras  when  th« 
numerator  and  denominator  arc  prime  to  each  other,  (Art. 
88,  Del'.  5);  as,  j|f  but  not  \. 

UK  VIEW. — 116.  What  is  the  effect. of  multiplying  both  terms  alike? 
117.  Of  dividing  both  terms  alike T  Illustrate  and  prove  those  propositions. 


78  RAY'S  HIGHER  ARITHMETIC. 


Rri.B. — Divide  both  terms  by  any  common  factor,  do  the  *ami 
to  ihe.  resulting  fraction,  and  so  on,  till  both  terms  are  prime  to 
eack  otlier. 

Or,  divide  both  terms  of  the  fraction  by  their  greatest  common 
divisor. 

licduce  1 8  to  its  lowest  terms. 

SOLUTION. — Dividing  both  terms 

bv  the  common  factor  2,  the  result  is     o  ~\  3  ft f^lfi 2    A 

If ;  dividing  this  by  5,  the  result  ia     1/»l 
|,  which  can  not  be  reduced  lower. 

Or,  dividing  at  once   by  10,  the  2          2 


V  »  j         UtViVAIU£        *»t<        VUVV          MJ          A  V  j         M*  V  «    »  f. 

greatest    common    divisor    of    both      J-"^S3==3 
terms,  the  result  is  f ,  as  before. 

DEMONSTRATION. — The  value  of  the  fraction  ia  not  changed, 
because  both  terms  are  divided  by  tho  same  number  (Art.  117). 

REDUCE  TO   THEIR  LOWEST  TERMS, 

ANS.  ASS.  ANS. 


1        3ft  2 

1.     45  •      -      * 

2X2  4 

•I*  .      .      7 


34   i  • 

•     T60       •        «        t 


T7J 


4.  -IS!    . 

5.  MI  .  .  n 

6.  iSf    .    .    {* 


7.  ffl 

8.  HI 

9.  \W 


EXPRESS  IN  THEIR  SIMPLEST  FORMS 


10.  923-5-1491      =M 

11.  890-4-1691     =i8 


12.  2261-^4123     = 

13.  C160-7-40480  = 


CASE  II. 

TO  BXDUCX  AN  IMPROPER  FRACTION  TO  A  WHOLE  OB 
MIXED  NUMBER. 

ART.  120.    RCLK. —  IHeide  the   numerator  by   the,   denomt- 
nalor ;  the  quotient  will  be  the  whole  or  mixed  number. 

NOTE. — If  there  be  a  fraction  in  the  answer,  reduce  it  t    tta 
lowest  terms. 

To  reduce  V  of  a  dollar  to  dollars,  divide  13  by    & 

eiaking  2$  dollars. 

DEMONSTRATION. — Since  6  fiftht  make  1  dollar,  there  wi'tl  be  na 
many  dollar?  in  13  fifths  as  5  fifths  are  contained  times  in  13  fifth? 
thai  is,  '2 \  dollars;  and  so  in  all  such  cases. 

REVIEW. — 118.   What   is  Reduction  of  Fractions?     119.    When  u  a 
fraction  in  iu  lowest  terms  f 


COMMON  FRACTIONS.  79 


1.  In    V    of  a  dollar,    how  many  dollars?  Ant.    4  I 

2.  In   'I7  of  a  bushel,  how  many  bushels ?  Am.  34 j 

3.  In  V<f  of  an  hour,  how  many  hours?     An*.  13-nz 


7.  'j?2  .     .  ^w.  105T7T 

8.  4f!°  .    .  An*.  327/3 

9.  1S37T80. 


REDUCE  TO  WHOLE  OR  MIXED  NUMBERS, 

4.      II  .     -4ns.      1. 

6.    *!$*     ,     .     Ans.    35. 
6.      B       .    .     Ans.  88 g 

CASE  III. 

TO    REDUCE    A   WHOLE    OR    MIXED    NUMBER    TO    AN 
IMPROPER    FRACTION. 

ART.  121.  RULE. — Multiply  together  the  whole  number  and 
the  denominator  of  the  fraction;  to  the  product  add  the  nu- 
merator, and  write  the  sum  over  the  denominator. 

Reduce  3 1  to  an  improper  fraction  ;  to  fourths. 

DEMONSTRATION. — In  1  (unit),  05 

there  are  4  fourths;  in  3  (units), 
there  are  3  times  4  fourths,  =  12 

fourths:  and  12  fourths -f- 3  fourths        12  =  fourths  in  3 
=  15  fourths.  3  =  fourths  in  fraction. 

REMARK. — 1.  This  dcmonstra-        15=fourths  in  3| 
tion  shows  that  the  whole  number  Ant.   V5 

is  really  the  multiplier,  and  the 

denominator  the  multiplicand  ;  but,  multiplying  by  the  denominator 
is  more  convenient,  and  gives  the  same  result,  (Art.  47). 

2.  This,  and  the  preceding  cn-c,  are  the  reverse  of,  and  mutually 
prove  each  other. 

1.  In  $71,  how  many  8ths  of  a  dollar?      .  Ans.    V 

2.  In  19|  gallons,  how  many  fourths?  .     .  Ans.    V 

3.  In  13| o  hours,  how  many  sixtieths?     .  Ans.  ^o7 

REDUCE   TO   IMPROPER   FRACTIONS, 

109T95  .     .    Ans.  2?f° 
5l?i     .     •    Ans.  'sYr2 


4    lit     . 

.     .    Ans.  V 

7. 

6.   15T8T  . 

.  Ans.  VV 

8. 

6.   127}^ 

.      Ans.  2H° 

9. 

.— 119.  What  is  the  rale  for  reducing  a  fraction  to  its  lowtft 
terms?  Prtve  the  rale.  120.  Give  the  rule  for  reducing  an  impropM 
frastion  to  a  whole  or  mixed  number.  Prove  it 


BO  RAY'S   HIGHER   ARITHMETIC. 

AHT.  122.  To  reduce  a  whole  number  to  a  fraction 
having  a  given  denominator,  is  merely  an  example  of  the 
preceding  case,  the  numerator  of  the  fractional  part  being 
zero,  (0).  It  is  done  by  multiplying  together  the  whol* 
number  and  the  denominator,  and  writin0  *he  product  ove) 
the  dc  nominator. 

To  reduce  4  to   a  fraction  whose   denominator   is  5,  v 
be  eame  as  to  reduce  4^  to  an  improper  fraction. 

1.  Reduce     7  to  4ths  ........     Ans.    V 

2.  Reduce  N  9  to  sevenths  ......     Ans.    V 

3.  Reduce  23  to  twenty-thirds  .....     Ant.  V/ 

4.  Reduce  19  to  a  fraction  whose  denominator  is  29. 

Ans.  W 
CASE  IV. 
TO  REDUCE  COMPOUND  TO  SIMPLE  FRACTIONS. 

ART.  123.  RCLE.  —  Multiply  all  the  numerators  together  foi 
a  new  numerator,  and  all  ike  denominators  together  for  a  new 
denominator. 

Reduce  I  of  ?  to  a  simple  fraction. 

iof  *  =  i^  =  A  Ans. 


DEMONSTRATION'.  —  f  of  §  =  2  times  ^  of  4  =2  limes  3  of  J  of  4, 
(since  5  of  4  is  the  same  as  »,  by  Art.  1O>,)=2  times  y'j  of  4, 
(since  |  of  £  is  the  same  as  7*5,  by  Art.  Ill,)  =  2  times  y4j, 
(since  T's  of  4  is  the  same  as  T45,  by  Art.  10G,)  =  785  (since  multi- 
plying the  numerator  multiplies  the  fraction,  by  Art.  112.) 

NOTE.  —  Before  applying  the  rule,  reduce  mixed  or  whole  num- 
bers to  a  fractional  form. 

1.  Reduce  i  of  f  of  3  7  to  a  simple  fraction. 
SOLUTION.  —  S|  =  2^2,  and  \  of  §  of  273  =  T404?  Ans. 

2.  |  of  f\  to  a  simple  fraction  .....    Ans.    \i 

3.  i  of  |  of  2f  to  a  simple  fraction.     .     .    Ans.  {!§ 

4.  |  of  2  of  85  to  a  simple  fraction.    .     .  Ans.  l^o 

NOTE.  —  Equal  factors  may  be  canceled  out  of  the  numerator  aril 
demmiiKitor,  us  out  of  any  other  dividend  and  divisor,  (Art.  1U1) 
before  the  multiplications  are  performed. 

REVIEW.  —  121.  Give  tho  rule  for  reducing  a  whole  or  raised  number  t« 
»n  improper  fraction.  Provo  Ik  Which  number  is  really  ihe  multiplier? 


COMMON   FRACTIONS. 


5.    Reduce  f  of  i9o  of  T7J  to  a  simple  fraction. 

S  o  L  r  T  i  o  s  .  —  The  factors  2,  3,  a 

ami  3  are  common  to  both  terms.      2X9x7 
Canceliug  them,  and  multiplying     -=r       |-v-       j-^  =  /o  Ans. 
together   the   remaining    factors,      P        Is         r£ 
the  result  is  7  twentieths. 

REDUCE   TO  SIMPLE   FRACTIONS, 

6.     \  of  |  of  4.    .Ans.  I    8.     lofji  of  21.  Am.    2, 


7.     i  of  \  of  21.  Ans. 


9.     £  of  |   of  3i  -4ns. 


10.  -|  of  I  of  4  of  8|  ........     Ans.  3] 

11.  i  of  I  of  ?  of  I  of  41  ......     Ans.     | 

12.  78T  of  f  of  T4?  of  IJ  of  Ti  .....     Ans.  112 

13.  If  of  79sof  isof  2?of  IsV      .     .     .     Ans.  & 
NOTK.  —  To  reduce  complex  to  simple  fractions,  see  Art.  132. 

CASE  V. 

ART.  124.  To  reduce  fractions  of  different  denomina- 
tors to  equivalent  fractions  of  a  common  denominator, 

RULE.  —  Multiply  both  terms  of  each  fraction  by  the  product 
of  alt  the  denominators  except  its  own. 

DK  M  O.VSTR  ATION.  —  Multiplying  both  terms  of  each  fraction  by 
the  same  number,  does  not  alter  its  value;  the  new  denominator  of 
each  fraction  will  be  the  tame,  since  it  will  be  the  product  of  the 
same  numbers;  viz.,  of  all  the  denominators. 

NoTE.-Reduce  compound  to  simple  fractions,  and  whole  or  mixed 
numbers  to  improper  fractious,  before  applying  the  rule. 

Reduce  2,  i  and  f  to  a  common  denominator. 

1  X  3  X  4  _  Ij^ncw  num. 

Both  ierms  of  the  first  frac-      'n  ^,  n  ^,   t  —  o~T 

...  ,.,,„...      JXoX-1       Z4  new  dcnom. 
lion  are  multiplied  by  3  X  4 

=  12;  of  the  second,  by  2  X     2^X  2  X  4  _  l_6^new  num. 
4  =  8^  and  of  the  third,  by  2     3  x  o  X  4  ~  2  4  new  dcnom. 

3^X  2  X  3  _  18  new  num. 
4x2x3~24new  denom. 

Since  the  denominator  of  each  new  fraction  is  the  product  of  the 
lame  numbers;  TIL,  all  the  denominators  of  the  given  fraction?,  it 

REVIKW.  —  122.  How  is  n  whole  number  reduced  to  a  fraction  having  • 
given  denominator?  12:}.  lluw  are  compound  fractions  reduced  to  siuipl* 
ones  ?  1'rovo  the  rule. 


IlAi  B   HIGHER   ARITHMETIC. 


is  unnecessary  to  find  this  product  more  than  once.     The  operation 
la  generally  performed  as  in  the  following  example. 

Reduce  2,  |  and  f  to  a  common  denominator. 
1X5XI7  =  35   1st  num. 

3X  2X  7  =  42  2d  num.  .        1§    42    «o 

6  X  2  X  5  =  60  3d  num.  A™'  ™»  "'  fil 

2X  OX  7  =  70  com.  denom. 

REDUCE   TO   A  COMMON   DENOMINATOR, 

1.  if,  I 

2.  i,  i,  i 

3.  I,  4,  I  ........  Am.  HI,  /A,  W 

4.  i,  *,  f,  I   .....    4u.  fig,  HI  Ml,  IIS 

5.  f,  iof8i,|of  f    .     .    .     .      Am.U,  W.SJ 

6.  !|i 


ART.  125.  When  the  terms  of  the  fractions  are  small, 
and  one  denominator  is  a  multiple  of  the  others,  reduce 
the  fractions  to  a  common  denominator,  by  multiplying 
both  terms  of  each  by  such  a  number  as  will  render  its 
denominator  the  same  as  the  largest  denominator.  Thit 
number  will  be  found  by  dividing  the  largest  denominator 
by  the  denominator  of  the  fraction  to  be  reduced. 

Reduce  ^  and  |  to  a  common  denominator. 

1x2      2 

SOLUTION  —  The   largest  denominator,  6,  is  a  —  =— 

multiple  of  3;    therefore,   if  we  multiply  both     3x2        b 
terms  of  g  by  6  divided  by  3,  which  is  2,  it  is      5  5 

reduced  lj  |.  7T          =~77 


1. 

2. 
3. 

REDUCE   TO   A   COMMON   DENOMINAT 

OR, 

tns.  I,  |,  g 

8      10      7 
T2,  T2,  T5 

llj  20)  20 

3,  |  and  iV 
I,  1,  T9o  and 

u.  .  . 

R  E  v  IE  w.  —  123.  If  there  are  any  mixed  numbers,  what  must  be  doneT 
What  may  bo  done  before  multiplying  ?  124.  How  are..fractions  of  dif- 
ferent denominators  brought  to  a  common  denominator?  Prore  the  rule. 
What  inust  be  done  with  compound  fractions  ?  With  whole  or  mixed 
numbers  ? 


COMMON  FRACTIONS. 


CASE  VI. 

ART.  126.  To  reduce  fractions  of  different  denomina- 
tors, to  equivalent  fractions  of  the  least  common  de- 
nominator. 

RULE. — Find  the  hast  common  multiple  of  the  given  denomi- 
nators; multiply  both  terms  of  each  fraction  by  the  quotient 
obtained  by  dividing  this  least  common  multiple  by  the  denomi- 
nator of  the  fraction. 

Reduce  3,  |  and  |  to  the  least  common  denominator. 
2)2    4    6      2)12      4)12      6)12 
236  3  2 

0v9v3  —  19        Jt*§ fi,     ^X3 9,     §X2_JO 

JX-XO  =  1^       2  x  «  =  i  2,    4X3=1 2,    8  x  Z  =  I  2 

least  com.  mul.  AM.  •&,  -&  and  j£ 

DEMONSTRATION. — Since  multiplying  both  terms  of  a  fraction 
by  the  same  number  does  not  alter  its  value,  (Art.  110),  each  of  the 
given  fractions  may  be  reduced  to  an  equivalent  fraction,  whose 
denominator  ia  any  multiple  of  its  own;  and  they  may  all  be  re- 
duced to  equivalent  fractions  of  the  least  common  denominator,  by 
taking  for  that  denominator  the  leatt  common  multiple  of  the  given 
denominators. 

After  getting  the  least  common  denominator,  both  terms  of  each 
fraction  are  multiplied  by  the  quotient  of  the  least  common  denomi- 
nator divided  by  its  own  denominator,  as  in  Art.  125. 

NOTES. — 1.  Before  commencing,  each  fraction  must  be  in  its 
Lowest  terms. 

2.  Reduce  compound  to  simple  fractions,  and  whole  or  mixed 
numbers  to  improper  fractions. 

3.  When  the  pupil  is  acquainted  with  the  principles  of  the  opera- 
tion, the  multiplication  of  the  denominators  may  be  omitted,  as  the 
new  denominator  of  each  fraction  will  be  equal  to  the  least  common 
multiple. 

4.  The  object  of  reducing  fractions  to  a  common  denominator,  is 
to  prepare  them  for  Addition  or  Subtraction. 

REVIEW. — 125.  When  one  denominator  is  a  multiple  of  the  others, 
bow  can  the  fractions  be  reduced  to  a  common  denominator?  126.  How 
are  fractions  of  different  denominators  reduced  to  a  least  common  de- 
nominator? What  must  be  done  if  a  fraction  is  not  in  its  lowest  terms? 
If  there  are  compound  fractions  ?  If  there  are  whole  or  mixed  numbers  T 
What  will  the  least  common  denominator  be? 


84  RAY'S   HIGHER   ARITHMETIC. 

REDUCE   TO    LEAST   COMMON    DENOMINATOR, 
1.        ,  43,  §       .........   An,.  V,,  T 


2.  i,  j|,  T9o    |     ...... 

3.  ?,|,  -|{ 

4.  i,  1'iVis* 

5.  ?>  T-2>  20'  Iff       .....        AllS.  eo»  GO>  tiO)  uo 

K  ,2       fi       3   nf    1434                                      /<   ,0     -  0      I  »     -'  I      I  7 

O.  3,    JO,    4    U1     75,    6<J           ...          AllS.    30'    30'    30'    30 

7.  143>  3j  and  73o  of  3}       ...    ^««.  W.  W.  2S 
*  See  Note  1,  preceding  page. 


ADDITION  OF  FRACTIONS. 

ART.  127.  Addition  of  Fractions  is  the  process  of 
adding  together  two  or  more  fractional  numbers. 

RULE. — Reduce  the  fractions  to  a  common  denominator;  add 
their  numerators  together,  and  place  the  sum  ocer  the  common 
denominator. 


8    (12 


24 

4 

0 

x 

3  = 

1 

^  ^ 

8 
12 

3 

2 

X 
X 

5  = 

7  = 

1 
l 

1 

new 

numerators. 

o 

2  x"l2  =  24 

least  com.  mul.  24  —  lil  A  us. 

DEMONSTRATION. — When  the  denominators  of  two  or  more  frac- 
tions are  the  same,  they  express  parts  of  the  same  size;  we  can,  there- 
fore,  add  1  four(h  ^  cenL 

2  fourths  >  as  we  would  add  •!  2  cents. 

3  fourths  )  (.  3  cents. 

The  sum  being  6  fourths,  (f),  in  one  case,  and  0  cents  in  the  other. 

That  is,  to  add  fractions  having  a  common  denominator,  find  the  yum 
of  the  numerators  and  write  the  result  over  the  denominators. 

l?ut,  if  the  denominators  are  different,  fractions  do  not  expresi 
things  of  the  same  unit  value,  and  can  not,  therefore,  be  added  to- 
gether (Art.  40);  in  the  preceding  example  we  can  not  add  fourths, 
eighth*  nnd  twelfth*,  but,  by  reducing  them  to  tu-cnty-ftmrth*,  they 
express  things  of  the  same  denomination,  nnd  can  then  be  added. 

R  R  v  i  K.  \v.  — 127.  Why  are  fractions  reduced  to  n  cnniin<>n  dviiuuiiuuuir? 
What  is  Addition  of  Fractions?  What  is  the  rule?  1'ruve  it. 


SUBTRACTION   OF   FRACTIONS.  H5 

NOTES.  —  1.  Before  commencing  the  operation,  each  fraction 
shoul'l  be  in  its  lowest  terms,  and  compound  fractions  must  be 
reduced  to  simple  ones. 

"2.    Mi  led    numbers   may  be  reduced  to  improper  fractions,  and 
then  H.Mfd;    or,  the  fractions  ma;    be  added,  and  then   the   who!« 
i» a :u U.- PI,  and  the  results  united. 
8.  After  ad  ling,  reduce  the  result  to  its  lowest  terms. 
l.^Add  iV  A.  1*.  iUnd  1!      .     .     .       Ans.        2. 
2f  43,  IMS  and  $3       ....       AM.      I\ 

3.  B'  I  and  T7ff        '       Ans.    1  \  % 

4.  5»  9  and  r2         Ans.        il 

5.  i>  l>  9  and  i4s An*.     2\s 

6.  la  and  22 Ans.     4^ 

7.  21,  31  and  41 Ans.  1011 

849i8jil  A  y  -i 

•  81  lot  3o  and  L\ Ans.     Ofj 

9.  T6o,  iV  T85  and  2j       ....       Ans.     3;f  5 

10.  1|,  21,  31  and  41      ....      Am.  11<U 

11.  I  of  j,  and  f  of  I  of  21      .     .      Ans.  1  rVo 

12.  What  is  i  + A +  4? Ans.  !TVs 

14.  /^  +  ll  +  t+H?      ....      ^liw.    4Jl 

15.  2" 

16.  4 

17.  I 

18.  i  of  61  +  T8s  of  ^  of  7J?    .     •      Am.    4/5 

19.  4  of  961+ f  of  U  of  51?     .      ^;w.  5911 

20.  ^  i 

SUBTRACTION  OF  FRACTIONS. 

ART.  128.  Subtraction  of  Fractions  is  the  process  of 
Indiug  tho  difference  between  two  fractional  numbers. 

RruE. — Reduce  the  fractions  to  a  common  denominator  ;  fnd 
Ihe  diflerenct  of  tticir  numerators,  and  place  it  over  the  commor 
denominator. 

R  B  VIK  w. — 127.  Before  commencing,  what  (bouM  be  done?  What  with 
mixed  number*  f  What  should  be  done  with  the  answer,  when  obuinodl 


SO  RAY'S   HIGHER   ARITHMETIC. 


Find  the  difference  between  f  and  7*5. 

SOLUTION.  —  The  fractions,  when  reduced  to  the  least  commoii 
denominator,  become  -ijg  nad  3  Si  their  difference  is  -390  =  730. 

DEMONSTRATION.  —  When  the  denominators  of  two  fractions  are 
ixe  sumo,  they  express  parts  of  the  same  size,  and  their  difference 
an  be  fonnd  as  in  the  case  of  whole  numbers. 

Thus  5  sevenths,  6  cents. 

3  sevenths,  3  cents. 

Difference,  2  sevenths  (|)  in  one  case,  and  2  cents  in  the  other. 

Cut,  if  the  denominators  are  different,  the  fractions  do  not  ex 
press  things  of  the  same  kind  ;  therefore,  one  can  not  be  subtracted 
from  the  other,  any  more  than  3  cents  can  be  taken  from  5  apples, 
(Art.  43);  in  the  preceding  example,  fifteen/  hs  can  not  be  taken  from 
six/Its;  but  by  reducing  them  both  to  thirtieths,  their  difference  can 
be  found. 

NOTES.  —  1.  Before  commencing,  reduce  compound  to  simple  frac- 
tions, and  see  that  each  fraction  is  in  its  lowest  terms. 
2.  After  subtracting,  reduce  the  result  to  its  lowest  terms. 


WHAT   IS  ANS. 


i.  I—  753?.  .  .  n 

2.     A-fVof^?  .  II 

311             3      ~f   5  9  173 

•      ~5i  —  v!5  01    5  f  .  757 


4.     T5r  —  73  of  4?  .     T 


K          11  4    o 

5.      71  —  S3  f     - 

f»  0  to  1  (I 

0.      55  —  15  f     .      .      .   76 

•77119 
7.       ?K  -  75  f 


763 


75  f     •      •       •    -235 


1089 

33  —  85f 


ART.  129.  Mixed  numbers  may  be  brought  to  improper 
fractions,  and  then  subtracted;  or,  the  fractions  may  be 
subtracted,  and  then  the  whole  numbers  and  the  results 
united.  In  the  latter  method,  if  the  lower  fraction  is 
live  larger,  increase  the  upper  fraction  by  the  number 
of  parts  in  1  unit,  and  carry  1  to  the  first  figure  of  whole 
numbers  in  the  lower  line. 

Thus,  6|     |  -f- 1  =  i 

6!-4!  =  V-y  =  V=l!;  or,_4}          !  =  l 

II 


RE vi  EW.— 128.  "What  is  Subtraction  of  Fractions?  What  is  the  rule? 
Prove  it.  Wh.it  should  be  done  before  commencing?  What  should  be 
dono  with  the  answer,  when  obtained?  129.  How  are  mixed  nutnben 
robtractfld  T 


MULTIPLICATION   OF   FRACTIONS. 


•WHAT   IS                              ANS. 

WHAT   IS 

AN9. 

9. 

12$—  1012?  .    Ill 

13. 

15  —  f?  .    .    . 

14* 

10. 

12;f  —  9sj?   .    32'« 

14. 

18  —  51?     .    . 

121 

11. 

5  5  1  —  2??.      .     3  iVi 

15. 

f  of  21  —  3}2? 

11 

12. 

7  7*5  —  34?  .    .    3}4 

16. 

3^—|  of  11? 

11 

V  of  44—  V  of  3£? 

13H 

18. 

Ans.  1( 

31  1  5 
ITJff 

19.  A  man  owned  at  of  a  ship,  and  sold  f  of  his  sharo 
how  much  had  he  left?  Ans.  fg 

20.  After  selling  4  of  f  +  i  of  f   of  a  farm,  what  part 
of  it  remains?  •  Ans.  H 


21.  3J  +  4I—  54  +  161—  741  +  10—  14i,    is  equal 
to  what?  A—   r-'1>J 


-  21  +  V  —  3 A  +  3^2  -f  8$  — 16]  =  what ? 

Ans.  .f| 

I  of  |  —  I  of  |  =  how  much  ? 


MULTIPLICATION    OF    FRACTIONS. 

ART.  130.  Multiplication  of  Fractions  is  the  process  of 
multiplication,  when  one  or  both  of  the  factors  are  frac- 
tional numbers.  It  embraces  three  operations  : 

1.  To  multiply  a  fraction  by  a  whole  number. 

2.  To  multiply  a  whole  number  by  a  fraction. 

3.  To  multiply  one  fraction  by  another. 

Since  any  whole  number  may  be  expressed  in  the  form 
of  a  fraction,  (Art.  108),  these  3  cases  may  all  be  per- 
formed by  this 

GENERAL  RULE  FOR  MULTIPLYING  FRACTIONS. 
Multiply  the  numerators  together  for  a  new  numerator,  and 
the  denominators  together  for  a  new  denominator. 

Multiply  I  by  |.      .     .     .     .     .     .      $  X  I  =  T8y  AM. 

DEMONSTRATION. — We  can  attach  no  other  idea  to  the  product 
of  4  fifths  by  2  thirds,  than  that  signified  by  2  thirds  of  4  fifths. 

REVIEW.— 130.  What  is  Multiplication  of  Fractions?  Wba!  does  it 
ttnbrace?  What  is  the  general  rule  ?  Prove  it. 


88  RAY'S  niOIIER  ARITHMETIC. 


But,  2  thirls  of  4  fifths  is  8  fifteenths,  (Art.  123);  therefore,  the  pro- 
luci  of  4  fifths  by  '2  thirds,  ia  also  8  fifteenths.     Hence,  these 

COROLLARIES. 

I  To  Mct.Tirt.r  A  FRACTION  nr  \  WHOLE  XUVBER. — Multiply 
the  n uine.ru lor  of  the  J 'ruction  by  the  whole  number,  and  wriU 
the  prinltn't  over  (he  denominator,  (Art.  11'2). 

Or,  lUcide  the  denominator  of  the  fraction  by  the  whole  n~im 
bttr,  irhen  it  can  be  done  withrmt  a  remainder,  and  over  ih» 
qnntir.nl  write  the  numerator,  (Art  115). 

II.  To  Mci.Tipi.r  A  WHOLE  XCMRER  DV  A  FRACTION. —  Mufti)  Jy 
the  irh»le  number  by  the  numerator  of  the  fraction,  and  dicide 
the  product  by  the  denominator. 

REMARKS. — 1.  After  indicating  the  operations,  if  the  numerator 
ami  denominator  contain  common  factors,  cancel  them  before  mul- 
tiplying. The  result  will  be  in  its  lowest  terms. 

2.  Multiplying  one  fraction  by  another  is  the  same  as  reducing  a 
compound  to  a  simple  fraction,  (Art.  123). 

EXAMPLES  FOR  PRACTICE. 
1.     12X12.     .  =9T33     4.    T9«X28.     .    =15f 


2.  lj  X  18  .     .  =    8{ 

3.  fix  24  .     .  =  14i 


5.    }i  X  30  .     .    =  20 
G.     33JX  5         .    =181 


SUGGESTION. — In  multiplying  a  mixed  number  by  a  whole  num- 
ber, multiply  the  fraction  and  the  whole  number  separately,  and 
add  the  products;  or,  reduce  the  mixed  number  to  an  improper  frac- 
tion, and  multiply  it;  as, 

|X5  =  31,    and  3X5  =  15;    and  15  -f  3]  =  181. 
Or,  3?  =  V,  and  V  X  5  =V  =  181. 

7.  45  X?    .    .  =35.    11.    28x32    .    =1021 

8.  50xii.    .  =39*    12.    4iX77«    .     .    =23o 

9.  25  X  I     .    .  =18!    13.     l^xM    .    .    =7V 
10.  32X2|.    .=76.     14.    Six!!    ....—II 
15.  What  will  3j  yards  of  cloth  cost  at  $4j  per  yard? 

Review. — 130.  How  i*  a  fraction  multiplied  by  a  whole  nwnber?  How 
Is  n  whulc  number  multiplied  by  a  fraction?  How  may  the  work  be  short- 
ened? Multiplying  fractions  is  equivalent  to  what  case  of  reduction! 
(low  may  a  mixed  nuuibur  be  multiplied  by  a  whole  number? 


DIVISION    OF  FRACTIONS. 


SDOOESTION. — In  finding  the  product  of  two  raized  numbers,  It 
is  generally  best  to  reduce  them  to  improper  fractions    thus, 


4t 
The  operation  may  be  performed  without  reducing  *>  * 

to  improper  fractions;  thus,  3  yards  will  cost  $13,|,          1  3 3 
and  ^  of  a  yard  will  cost  ^  of  $4^  =  $!^;  hence,  1 1 

the  whole  will  cost  $15.  -te 


18.  121  X  3T3r  .   =402 

19.  TUX  3T7g  .   =26l 


16.    61  X  4h  .    .  =30. 
17     4»X2'|  .    .  =12| 

20.  Multiply  £  of  8   by   \  of  10   .     .     .     .   Ans.  4. 

21.  Multiply  f  of  5|   by    f  of  3i       ...   Ans.  Y£ 

22.  Multiply  f  off  of5|   by   f  of  31    .     .   ^ns.  ±& 

23.  Multiply  5,  41,  21    and    f  of4|     .     .   AM.  M 

24.  Multiply  |,  |,   T6T,  i  of2i,   t  of  3}     .   Ans.  /ft 

25.  Multiply  f,   ?,   T9T,   3s  and  3^    ...      Ans.    4. 
.  Multiply  3i,  4^,  51,  f  of  h,  6|       .     AM.  49. 

27.  At  g  of  a  dollar  per  yard,  what  will  25  yards  of 
cloth  cost?  Ans.  $21 3 

28.  A  quantity  of  provisions  will  last  25  men  12.(  days: 
how  long  will  the  same  last  one  man?     Ai^.  318]  days. 

29.  At  3j  cents  a  yard,  what  will   2*   yards  of  tape 
cost?  Ans.  9jj  cts. 

30.  What  must  be  paid  for  f  of  |  of  a  lot  of  ground 
that  cost  §184  ?  Ans.    $7a; 

31.  K  owns  |  of  a  ship,  and  sells  |  of  his  share  to  L: 
what  part  has  he  left?  Ans.  35 

32.  B  bought  f  of  a  farm  of  21 9|  acres,  and  sold  i  of 
his  part  to  C:  what  part  of  the  whole,  and  how  many  acres, 
did  he  sell?  Ans.  7*5,  and  29|  acres. 


DIVISION   OF  FRACTIONS. 

ART.  131.  Division  of  Fractions  is  the  process  of  divi- 
sion, (Art.  5S),  when  the  dividend  or  divisor,  or  both,  ar« 
fractional  numbers.  It  embraces  three  operations : 

1.  To  divide  a  fraction  by  a  whole  number. 

2.  To  divide  a  whole  number  by  a  fraction. 

3.  To  divide  one  fraction  by  another. 

8 


90  RAY'S   HIGHER  ARITHMETIC. 

Since  any  whole  number  may  bo  expressed  in  the  fora 
of  a  fraction,  (Art.  108),  these  3  eases  may  all  bo  per- 
formed by  this 

GENERAL  RULE  FOR  DIVIDING  FRACTIONS. 

Invert  the  divisor;  then  multiply  the  numerators  together  fot 
•  new  numerator,  and  the  denominators  for  a  new  denominator. 

Divide  I  by  |. 

|  inverted  =  f,  and  |  X  f  =  f  =  lg  Ans. 

1st  DEMONSTRATION. — Suppose  the  divisor  were  2  instead  of  |; 
the  quotient  would  be  |xi==|i  (Art.  114).  But,  the  real  divisor, 
(j),  is  only  one  third  as  large  as  the  supposed  divisor  (2);  there- 
fore, the  real  quotient  must  be  three  times  as  large  as  the  supposed 
quotient,  (Art.  73),  and  3  times  |  =  §,  (Art.  112);  which  agrees 
with  the  rule. 

2d  DEM. — It  hag  been  shown,  (Art.  GO),  that  the  divisor  and  divi- 
dend must  be  of  the  same  denomination;  hence,  to  find  how  often 
2  thirds  is  contained  in  3  fourths,  reduce  them  to  ttcelftht.  But, 
§  =  -/TV,  and  |  =  fr,  and  8  twelfths  in  9  twelfths  is  the  same  as  8 
in  9;  that  is,  g  =  lg  times. 

Hence,  the  mlo  might  have  been  expressed  thus : 

To  divide  one  fraction  by  another,  reduce  them  both  to  a  com 
mon  denominator,  and  divide  the  numerator  of  the  dividend  by 
the  numerator  of  the  divisor. 

COROLLARIES. 

I.  To  DIVIDE  A  FRACTION  BT  A  WHOLE  NUMBER. — Multiply  the 
denominator  of  the  fraction   by  the  whole  number,  and  over  the 
product  write  the  numerator. 

Or,  Divide  the  numerator  of  the  fraction  by  the  whole  number, 
when  it  can  be  done  without  a  remainder,  and  under  the  quotient 
write  the  denominator. 

II.  To  DIVIDE  A  WHOLE  NUMBER  BV  A  FRACTION. — Multiply  the 
nhole  number  by  the  denominator  of  the  fraction,  and  divide  the 
product  by  the  numerator. 

REVIEW. — 130.  How  may  mixed  numbers  be  multiplied  together? 
131.  What  is  Division  of  Fractions?  What  does  it  embrace?  What  is 
the  general  rule  ?  Illustrate  it.  What  is  the  1st  demonstration?  <he  2d? 
[low  is  a  fraction  divided  by  a  whole  number  ?  How  is  a  whole  number 
divided  by  a  fraction  ? 


DIVISION  OF   FRACTIONS. 


Tho  rule  may  be  simply  demonstrated,  as  follows: 

8d  DEM. — Inverting  the  divisor,  shows  how  often  it  is  contained 
in  a  unit,  and  (his  multiplied  by  the  dividend,  shows  how  often  it 
contains  the  divisor;  in  the  example  given,  since  3  is  contained 
in  a  unit  3  times,  f  is  contained  in  a  unit  ^  times,  (Art.  74),  and 
in  |,  it  is  contained  |  X  f -  =  f  times,  (Art.  130). 

NOTES. — 1.  Before  commencing,  reduce  compound  to  simple  frac- 
tions, and  mixed  numbers  to  improper  fractions;  also,  express  whol* 
numbers  in  the  form  of  fractions. 

2.  In  all  cases,  reduce  the  result  to  its  lowest  terms. 


EXAMPLES  FOR  PRACTICE 


1. 

79ff- 

-3 

3 

•     •     .    =  7  a 

8. 

21 

25- 

1  4 
-7?    • 

2. 

23- 

-7 

2 
•        •        •        =  23 

9. 

3S- 

-1?    • 

3. 

f  _ 

-8 

.      .      .      =  43T5 

10: 

11- 

-5    . 

4. 

6- 

2 
-3 

.    .    .    =9. 

11. 

81 
8  — 

-3 

5. 

21- 

-790 

.    .    .  =  23J 

12. 

19i-f-ll 

6. 

3- 

-i 

.    .    .  =   1  i 

13. 

73i_i_0^ 

7. 

3  - 

-4\j 

.    .    .  =  26j    14. 

5411  -i-2£ 

15. 

Divide 

U  by  i  of  |  of  7z 

. 

7 
=  HO 


.     =101 
i.    =  24 

.     Ans.  | 
10.     Divide  T3ff  of  T3a  of  TS2  by  37j  of  il   .     .   ^HS.  gf 

NOTE. — After  indicating  the  operations,  if  the  numerator  and 
denominator  contain  common  factors,  cancel  them  before  multiply- 
ing. The  result  will  be  in  its  lowest  terms. 

17.     Divide  1  of  2^  by  I  of  1 1 

SOLUTION. — 2s  =1,  and  If  =  s 

2 

2 


$     3    $    y 

SUGGESTION. — Tn  the  division  of  fractions,  or  other 
operations  by  cancellation,  it  is  often  convenient  to 
place  the  divisors  on  the  left  of  a  vertical  line,  and  the 
multipliers  on  the  right. 


REVIEW. — 131.  What  is  the  3d  demonstration?     Before  commencing 
irh.it  must  be  done?     How  may  the  work  be  shortened  ? 


92  RAV&  HIGHER   ARITHMETIC. 

18.  Divide  |  of  $  by  !  of  S Ans.  2| 

19.  Divide  I  of  3|  by  if  of  7 Ant.  gf 

20.  Divide   '  of  |  X  a¥q  by  fs  of  31       .     .   Ant.     I 

21.  Divide  ?of5A  by  §  of  A  of  3]    .     .     .  Ans.  1| 

22.  Divide  3  of  2  Of  T4T  by  |  of  J  of  t      .     .  Am.  A 

23.  Divide  lg  times  4*  by  ITT  times  3«     .  ^ln«.  If 

24.  Divide  3?  by  |  of  8]  times  A  of  3  rso   .  Ans.     4 

25.  Divide  A  of  I  of  27*  by  -$  of  f V  of  5 -A  4/w.  34-J3 

26.  What  is  23  X  5  of  19^  H-43  X  j36  of  8?  ^HS.  3^ 
ART.  132.    To  reduce  complex  to  simple  fractions. 

RULE. — Divide  the  numerator  by  the  denominator  as  in  Divt> 
tion  of  fractions,  (Art.  131). 

Or,  Multiply  both  terms  of  the  complex  fractions  by  the  least 
common  multiple  of  the  denominators  of  their  fractional  parts. 

1  5 

Reduce  —  to  a  simple  fraction. 
24 

iu  111  2    o    ' 

OPERATION.  :  then = =  —  X—    =— 

9?  —  U'  '    fi    '    4.        a        TT         3 

«•  1   —     4  \)  rr          1)  Q     i  J.  O 

The  least  common  multiple  of  6  and  4  is  12;  hence, 

lgXl2      22      2 


DEMONSTRATION. — Since  every  fraction  indicates  that  the  nu- 
poerator  is  to  be  divided  by  the  denominator,  the  1st  Rule  needs 
no  demonstration. 

Since  the  value  of  a  complex  fraction  is  not  changed  when  both 
terms  are  multiplied  by  the  same  number,  (Art.  1113),  and  since 
12  is  tlic  most  convenient  multiplier  tli.it  will  cause  the  small 
fractious  to  disappear,  the  reason  of  the  2d  Rule  is  evident. 

REDUCE   TO    SIMPLE    FRACTIONS, 


,.i 

.  ^ln«.  ^3, 

3. 

4j 

.  .4ns.  If 

5. 

121 

Ans.  \l 

y 

24 

18 

o    *•* 

.  ^ns.  | 

4. 

2A 

.   ^!HS.  lg 

6. 

G2 

Ans.  3j 

SI 

2iV 

Wi 

REVIEW. — 132.  AVbat  arc  the  rules  for  reducing  a  complex  to  a  simple 
fraction?     Tllustr.itp  and  prove  thorn. 


DIVISION    OF   FRACTIONS. 


Complex  fractions  may  be  multiplied  or  divided,  bj 
reducing  them  to  simple  fractions.  The  operation  may 
often  bo  shortened  by  cancellation. 


7  1x1* 
10X8   ' 

8  —v  M. 


9     £*xi^? 


10.  -H    - 

2|       12i 


11.   " -i-  v    A*,  u 

401       73 

12.    2TJ     2T\      J/u  3 
21      8i70 


TO  FIND  THE  GREATEST  COM.  DIVISOR  OF  FRACTIONS. 

ART.  133.  RULE. — Reduce  the  numbers  to  simple  fraction*, 
and  to  their  lowest  terms ;  Jind  the  greatest  common  divisor  of 
the  enumerators,  and  divide  it  by  the  least  common  multiple  of 
the  ^nominators:  the  quotient  will  be  the  greatest  common  divisor 
of  the  fractions. 

NOTES. — 1.  If  the  numerators  are  prime  to  each  other,  and  the 
denominators  are  also  prime  to  each  other,  the  greatest  common 
divisor  of  the  numbers  will  be,  1  divided  by  the  product  of  the 
denominators. 

2.  The  greatest  common  divisor  of  more  than  two  fractions  can 
be  obtained  by  first  finding  the  greatest  common  divisor  of  two 
of  them,  then  of  this  and  a  third,  and  so  on ;  the  last  divisor  will 
be  the  greatest  common  divisor  of  all. 

Find  the  greatest  com.  divisor  of  261  and  1085f 

SOLUTION. — The  numbers  when  reduced  to  the  form  of  fractions, 
are  !J*  and  8^85;  the  greatest  common  divisor  of  the  numerators, 
105  and  8085,  is  15,  (Art.  95),  and  the  least  common  multiple  of 
the  denominators,  4  and  8,  is  8,  (Art.  99);  hence,  the  greatest 
common  divisor  of  the  given  numbers  is  'g*  =  lg 

DEMONSTRATION. — If  the  fractions,  '§5  and  8^64,  are  divided 
by  15  the  greatest  common  divisor  of  their  numerators,  the  quotients 
»re  fractious,  viz:  |  and  *|9.  If  the  divisor  15  be  divided  by  8, 
the  quotients  must  be  multiplied  by  8,  (Art.  73),  and  become  whole 
numbers,  viz:  14  and  579.  Now,  8  is  the  smallest  possible  number, 
which,  used  as  a  multiplier,  will  convert  the  quotients,  4  and  °|9, 


REVIEW. — 133.    What  is   the   rule  fur  finding   the   greatest   common 
divisor  of  fractions  ?     Illustrate  and  prove  it 


94  RAY'S   HIGHER    ARITHMETIC 


into  whole  numbers  at  the  same  time,  since  it  is  the  least  common 
multiple  of  the  denominators  4  and  8;  therefore,  as  'g%  when  used 
as  a  divisor,  gives  the  smallest  possible  whole  numbers  for  quo 
tients,  it  must  be  the  greatest  common  divisor  of  the  given  numbers 

FIND   THE   GREATEST   COMMON    DIVISOR 

1.  Of  83 J,  and  268.? An*.  2T* 

2.  Of  14T72  and  95| Ans.     & 

3.  Of  59J  and  735i| Ans.  2.}§ 

4  Of  23T7g  and  213^f An*.  2|i 

5.  Of  418!  and  17721 An*,     jf 

6.  Of  237!  and  1751-1 Ans.    5'i 

7.  Of  261}!  and  652-U An*.  4H 

8.  Of  44^,  546!  and  3160 An*.    4| 

9.  Of  137^,  4781  and  2093!  .  .  .  .  Ans.  Sk 
10.  Of  39774,  1022T7<j,  2954T75, 1680U  .An*.  1^75 

TO  FIND  THE  LEAST  COMMON  MULTIPLE  OF  FRACTIONS. 

ART.  134.  RCLE. — Reduce  the  numbers  to  simple  fraction 
and  to  their  loivcst  terms.  Find  the  least  common  multiple  of 
the  numerators,  and  divide  it  by  the  greatest  common  dirisor  of 
the  denominators;  the  quotient  will  be  the  least  common  mul- 
tiple of  the  fractions. 

Find  the  least  com.  multiple  of  3?,  4e,  lg,  iV 

SOLUTION. — The  numbers,  when  reduced  to  the  form  of  simple 
fractions,  become  'j5,  *g°,  f  and  f\ ;  the  least  common  multiple  of 
the  numerators  is  225;  the  greatest  common  divisor  of  their  denomi- 
nators is  2;  the  former  divided  by  the  latter  gives  112 A,  the  leas.* 
common  multiple  of  the  given  numbers. 

DEMONSTRATION. — If  the  given  numbers  were  the  whole  num- 
bers 15,  25,  9,  5,  their  least  common  multiple  would  be  225,  which 
Would  contain  them  respectively  15,  9,  25  and  45  times;  moreover, 
225  contains  Vj5,  285>  f,  fV,  respectively  4  X  l-r>,  6X9,  8x25, 
12X45  times,  since  dividing  the  divisor  multiplies  the  quotient, 
(Art.  73);  and  £  of  225  =  112.2  will  contain  the  same  numbers  half 
as  many  times  respectively,  viz:  2  X  15,  3X9,  4  X  25,  6  X  45 
times;  and  since  no  other  number  will  divide  all  these  quotients 
exactly,  112j  is  the  least  number  which  will  contain  the  fractions 
•xactly,  and  is,  therefore,  their  least  common  multiple. 


DIVISION  OF   FRACTIONS.  95 

HEM  ARK.  —  The  leaat  common  multiple  of  fractional  numbcri 
c>*n  also  be  found  by  Rule  3,  Art.  99,  taking  care  to  obtain  tht 
greatest  common  divisors  required,  by  the  rule  in  the  last  article. 

FIND  THE  LEAST  COMMON  MULTIPLE 

1.  Of  f  ,  S,  It,  §  and  $    ......       Ans.     60. 

2.  Of  4A,  6|,  58  and  10$  .....      Ans.  472$ 

3.  Of  3},  4|,  ft,  5$  and  12£     .    .     .      Ans.  350. 

4.  Of  14?,  OI'T,  16j  and  25       ...      Jns.  100. 

5.  Of  181,  CGI,  21|  and  15       .    .    .      Ans.  600. 

6.  Of  9!,  22  1A,  2575s  and  12?3ff       .     .  Ans.  1G70G[ 

7.  Of  84,  10§,  5[f,  6ft  and  f         .    .    4/«.  1437] 

8.  Of  ft,  T75,  US,  3  2*r  and  3U      .     .     ^ln*.  2245} 

9.  Of  121,  131,  143,  15i  and  17^    -  Ans.  11400. 

ART.  135.     PROMISCUOUS  EXERCISES. 

1.   Add  together  3i,  4],  5|,    f  of  |,   and  ^  of  3  of  |. 

ylns.  1332 


2.  The  sura  of  Ij'a  and  --j  is  equal  to  how  many  ti 
their  difference?  Ans.  5  times. 

3.  What  is  J2J+-  of  ~  —  1|  1-5-laYs?  ^,,s.  5. 

' 


4.  Reduce     -|^|r  5   and  ^  X  (100  -  ^  +  £)  to 

their  simplest  forms.  Ans.  16  and  26iV^ 

5.  What  is  J  of  5i  —  1  of  3g?  Ans.  35 

6.  What  is  |f  X  T8ft  X  3??  X  II  equal  to?       Ans.  T5g 

2~<L 

T8 


8. 

0 

i        i  ' 

Alan            V               .   \ 

3 

/    **               V  - 

J  —  1 

32 
(2  +  1)  ^-(8 

3 

5 

i,.,i  9 

(2-j)X(4-i 

REVIEW.— 134.   \Vhat  is  the  rule  for  finding  tho  lout  common   mol 
uple  of  fraotionn?     Illustrate  and  prove  it. 


96  RAY'S   HIGHER   ARITHMETIC. 

45x4]  x4S  —  1 

10.  —. =  what?          Ans.  4i« 

4]  X  4a  — 1 

11.  Add  |  of  43  of  |,   £  X  I  of  II,  and  \         Ans.  |§ 

12.  5  of  V  of  what  number,  diminished   by     ,  '^  7  > 

leaves  gj?  ~b"T930 

Ans.  /A 

13.  Find   the   least  common  multiple  of  the   numbers 
from  10  to  20  inclusive.  Ans.  232792560. 

14.  K  leaves  L  for  N,  (109  miles   apart)  at  the  same 
time  that  B  leaves  N  for  L.     K  travels  7^  miles  per  hour, 
and  B,  8|  miles  per  hour  :   in  how  many  hours  will  they 
meet,  and  how  far  will  each  have  traveled  ? 

Ans.  6ei  hours.     K,  51 A 'f  miles;  B,  57iV  miles. 

15.  What  number  multiplied   by  g  of  7    of  311    will 
produce  2i?  Ans.  1]  jj 

16.  What,  divided  by  If,  gives  14.??  Ans.  23  f 

17.  What,  added  to  I4g,  gives  29"?         Ans.  15rr|5 

18.  I   spend  |   of  my  income  in  board,  g  in  clothes, 
and  save  $60  a  year:  what  is  my  income?    Ans.  $216. 

19.  Find  the  Gr.  C.  D.  (greatest   common    divisor),  of 
96,  120,  160  a.nd  200,  and  the  least  com.  mul.  of  13,  19, 
57  and  65.  Ans.  8  and  3705. 

20.  The  least  com.  mul.  of  10,  24,  35  and  an  unknown 
number  prime  to  these  three,  is  9240.     What  is  the  un- 
known number?     (See  Rem.  1,  Art.  97).  Ans.  11. 

21.  What  two  numbers  between  35  and  840,  have  the 
former   for   their  G.  C.  D.,  and   the  latter   for   their  least 
com.  mul.?      (See  Note  3,  Art.  99).  Ans.  105  and  280. 

22.  Find  a  number  between  697  and  731,  which  shall 
have  with  each,  the  same  greatest  common  divisor  that  they 
have  with  each  other.  Ans.  714. 

23.  The   Gr.  C.  D.  of  three   numbers   is   15,  and   their 
least  com.  mul.  is  450.     What  are  the  numbers? 

^Ins.  30,  45  and  75. 

24.  2  is  what  part  of  3  ?  Aim.  I 

25.  Divide  I  of  3i    by  \\  of  7;  and  A  of  I  of  27'   by 
J  of  T3t  of  51  Ant.  II  and  34 U 

26.  Multiply  T7T  of  2i  by  T3s  of  191 ;   and  divide  I  of 
I  of  14  j    by  T3r  of  '|  of  13$.  Ans.  7^r  and  Oft 


DECIMAL   FRACTIONS.  97 


27.  Add  together  2i$,  3j,  2?  and  Si6!,  and  divide  th« 
sum  by  2|f.  Ans.  4rVTO 

I3  -J-2| 

28.  Express  -  -    as  a   simple   fraction;    and   aisu 

52  +  4i 

multiply  g  of  2.[  by  f  of  TV  Am.  §!§  and  /2 

.X29.  A  bequeathed  ^  o  of  his  estate  to  his  elder  son ;  the 
rest  to  his  younger,  who  received  $525  less  than  hia 
brother.  What  was  the  estate?  Ans.  $5250. 

30.  Find  the  sum,  difference,  and  product,  of  83    and 
2k  ;  also  the  quotient  of  their  sum  by  the  difference. 

Ans.  sum  673,  diff.  Iff,  prod.  8-rs,  quot.  3T6izV 

31.  A  cargo  is  worth  7  times  the  ship:  what  part  of  the 
cargo  is  fg  of  the  ship  and  cargo  ?  Ans.  -fa 

32.  If  a  railroad  car  runs  112f  miles  in  5i  hours,  what 
is  the  rate  per  hour?  Ans.  21 1^ 

33.  Subtract   f  of  ji  of  6|   from    g  of  51 ;    and  mul- 
tiply 19|  by  -  of  -|-  Ans.  383  and  5} 

5       o.» 

34.  65  is  what  part  of  10i7T?  and  reduce  to  its  sim- 
plest form  I  —  4  +  i— -If  Ans.  \i  and  sV* 

35.  Multiply  -;>  14|,  j>  ;->  r|»  and  6.       Ans.  135g 

5"g  4    7s  24 

36.  i  of  |  of  what  number  equals  9fl?       Ans.  20. 

37.  A    63   gallon   cask   is   |  full:    9h    gallons   being 
drawn  off,  how  full  will  it  be?  Ans.  |§| 

38.  If  a  person  going  3|  miles  per  hour,  performs  a 
journey  in  14|  hours,  how  long  would  he  be,  if  he  trav- 
eled 5f  miles  per  hour?  Ans.  10A§  hours. 

39.  A  man  buys  32|   pounds   of  coffee  at  I7g  cts.  a 
pound:  if  he  had  got  it  4§   cts.  a   pound  cheaper,   how 
many  more  pounds  would  he  have  received?  Ans.  ll^fi 


IX.   DECIMAL  FRACTIONS. 

ART.  136.  A  Decimal  fraction  derives  its  name  from 
the  Latin  word  decent,  meaning  ten,  and  is  so  called,  be- 
cause its  denominator  is  always  1  with  ciphers  annexed: 
being  10,  or  the  product  of  several  10's;  thus, 

37     84015        ,692  ,      .       ,   r       .. 

— >    -        and are  decimal  tractions. 

too    1000          1000000 

9 


08  RATS  HIGHER   ARITHMETIC. 


All  the  rules  and  operations  in  the  several  cases  «»? 
Common  Fractious  applj  as  well  to  Decimal  Fractions 
when  thus  written.  But,  a  more  simple  and  convenient 
Notation  ha?  been  devised  for  them  similar  to  that  of 
whole  numbers. 

This  notation    consists    in    writing  the    numerator,  and 
placing  a  point,  (.),  so  that  the  number  of  figures  on  the 
ight  of  it  shall    be   equal    to  the  number  of  ciphers   in 
ho  denomiLator. 

The  decimal  fractions  before  given,  when  expressed  in  this  way 
are  .37  and  84.015  and  .000002 

The  use  of  the  point  instead  of  the  denominator,  saves 
time  and  space,  while  no  mistake  is  likely  to  occur  on 
this  account,  since  the  denominator  can  be  obtained  a? 
follows : 

The  denominator  of  any  decimal  fraction  is  1  with  as 
many  ciphers  annexed  as  there  are  figures  on  the  right  of 
the  point. 

ART.  137.  A  decimal  fraction,  when  written  with  a 
point,  is  simply  called  a  decimal. 

The  places  and  figures  on  the  right  of  the  point  arc 
called  decimal  places  and  decimal  figures,  to  distinguish 
them  from  the  places  and  figures  of  whole  numbers. 

The  point,  (.),  is  called  the  decimal  point  or  tcpa.Ta.trix; 
it  separates  the  decimal  places  from  the  places  of  whole 
numbers. 

A  pure  decimal  "has  only  decimal  figures;  as,  .02319 

A  mixed  decimal  has  figures  of  whole  numbers;  a£, 
281.63 

A  complex  decimal  has  a  common  fraction  in  its  right- 
hand  place;  as,  .8i  and  2.622 

A  whole  number  may  be  regarded  as  a  decimal,  by  supposing  a 
point  to  be  on  the  right  of  its  units'  place;  as,  154  =  154. 

REVIEW. — 136.  What  is  a  Decimal  Fraction?  Why  BO  called?  Give 
•samples.  What  mode  of  expressing  them  has  boon  adopted?  Illustrate 
it  What  is  the  advantage  of  writing  decimal  fractions  with  a  point? 
When  thus  written,  how  can  the  denominator  be  known?  137.  What  if 
a  decimal  fraction  called,  when  written  with  a  point?  What  are  the 
figures  on  the  right  of  the  point  called?  What  is  the  point  called? 
What  is  a  pure  decimal?  What  is  a  mixed  decimal?  a  complex  decimal? 
Qiv»  examples.  How  may  every  whole  number  be  regarded  ? 


NUMERATION   OF   DECIMALS  «|<J 


NUMERATION  OF  DECIMALS. 

ART.  138.  Since  .6  =  -ft;  .06  =  Tgff;  and  .006  = 
r/fo^,  any  figure  expresses  tenths,  hundredths,  or  thoitsanths, 
according  as  it  is  in  the  1st,  2d,  or  3d  decimal  place 
hence,  these  places  are  named  respectively  the  tenths',  the 
\undretht-')  the  thousandths'  place;  other  places  are  named 
n  the  same  way,  as  seen  in  the 

TABLE  OF  DECIMAL  ORDERS. 


"cfioc^Sa83- 
Ss^Ss^SS- 

H  WH  H3*s  H  W£ 


1st  place  .2      ....  read  2  tenths. 

2d  ..  .08 8  Hundreths. 

3d  ..  .005       5  Thousanths. 

4th  ..  .000*7 *7  Ten -thousandths. 

5th  ..  .00003      .      .     ..  3  Hundred-thousandths. 

6th  ..  .000001.      .     ..  1  Milionth. 

7th  ..  .0000009     .     ..  9  Ten-Millionths. 

8th  ..  .00000004.     ..  4  Hundred-millionths. 

9th  ..  .000000006    ..  6  Billionths. 

The  names  of  the  decimal  orders  are  derived  from  the  names 
of  the  orders  of  whole  numbers.  The  table  may  therefore  be  ex- 
tended to  Trillionths,  Quadrillionths,  &c. 

ART.  139.  By  inspecting  the  table,  and  recollecting 
that  1  tenth  =  10  hundredths,  and  1  hundredth  =  10 
thousandths,  it  is  clear  that  the  decimal  places  decrease 
in  value  from  left  to  right,  like  the  places  of  whole  num- 
bers, and  according  to  the  same  law,  viz: 

1  in  any  place  equals  10  in  the  next  right-hand  place. 

This  law  holds  good  in  every  mixed  decimal,  like 
715.2309;  for,  in  all  such,  the  last  figure  of  whole  num- 
bers, (5),  is  units,  and  the  first  decimal  figure  on  the 
right  is  tenths,  and  1  unit  =  10  tenths. 

REVIEW. — 138.  What  is  tho  name  of  the  1st  decimal  place?  The  2dt 
Third?  Why?  Repeat  tho  Table  of  Decimal  Orders. 


100 


RAY'S   HIGHER   ARITHMETIC. 


ART.  140.  Since  decimals  are  subject  to  the  same  lav* 
of  local  value  as  whole  numbers,  like  them,  also,  they 
can  be  read  in  either  of  two  ways. 

RULE  FOR  READING  DECIMALS. 

1st.  Head  in  succession  the  value  of  (he  separate  figures  which 
compose  the  decimal;  or,  which  is  much  mure  convenient, 

2d.  Read  the  decimal  as  a  whole  number,  and  annex  Hie  name 
of  the  right-hand  place. 

The  decimal  .004038,  is  read,  4038  millionths. 

DEM. — The  reason  of  the  rule  depends  on  the  law  of  local  value, 
viz:  "1  in  any  place  equals  10  in  the  next  right-hand  place." 

Commencing  with  the  first  significant  figure,  4  of  the  3d  place 
equal  40  of  the  4th  place;  40  of  the  4th  place  equal  400  of  the 
6th  place,  which,  with  the  3  already  there,  make  403  of  the  oth 
place;  finally,  403  of  the  6th  place  equal  4030  of  the  6th  place, 
wnich,  with  the  8  already  there,  make  in  all  4038  of  the  Gth  place ; 
or,  4038  millionths,  since  the  Gth  place  expresses  millionths. 

A  mixed  decimal  may  be  read  altogether  as  a  decimal ; 
or,  the  whole  number  may  be  read,  then  the  decimal. 

Thus,  71.062  may  be  read  71062  thousandths;  or,  71  units,  and  62 
thousandths. 

EXAMPLES  TO  BE  READ. 


1. 

.9 

9.  00.100 

17.  41.14414 

2. 

.04 

10.  180.010 

18.  411.4414 

3. 

.305 

11.  20300.0 

19.  4.114414 

4. 

.7200 

12.  40.68031 

20.  15.0046J 

5. 

.5060 

13.  207.2007 

21.  73002.11 

6. 

1.008 

14.  .0900001 

22.  .000000^ 

7, 

9.00i 

15.  61.001001 

23.  .200006 

8 

105.02 

16.  9230010.0 

24.  526.000 

25.  12.3333333     28.  .43*7800629 

26.  643000.643     29.  1000000.0303 

27.  4009.62007     30.  200000.000006 

REVIEW. — 139.  How  do  decimal  places  resemble  those  of  whole  num- 
bere?  What  ia  the  law  that  governs  both  ?  Does  this  law  apply  to  mixed 
decimals?  Why?  140.  In  how  many  ways  may  decimals  be  read  ?  What 
la  the  1st  ?  The  2d  ?  Prove  the  2d  role.  How  may  mixed  decimals  be 
road?  flive  an  example. 


NOTATION    OF   DECIMALS. 


101 


NOTATION   OF   DECIMALS. 

ART.  141.  Write,  eighty-three  thousand  and  one  bil« 
lionths. 

DEMONSTRATION. — The      1st  step  83001  =  Numerator. 
number  of  parts   must   bo     2d    step  .000083001  =  Decimal, 
written  as  a  whole  number, 

(Art.  136) ;  the  right-hand  figure  must  express  parts  of  the  given 
size,  (See  last  Rule) ;  hence,  the 

RULE  FOR  WRITING   DECIMALS. 

Write  the  numerator;  fix  the  point  so  that  the  right-hand 
figure  shall  be  of  the  same  name  as  the  decimal. 

REMARKS. — 1.  In  fixing  the  point,  it  may  be  necessarj  to  prefix 
ciphers  to  the  numerator,  as  in  the  example  just  given;  but  in  the 
case  of  an  improper  decimal  fraction,  the  point  will  fall  between  two 
of  the  figures ;  thus,  346  tenths  is  written  34.6,  which  may  also  be 
read  34  units  and  6  tenths. 

2.  The  operations  under  this  and  the  preceding  rule  se^ve  to 
prove  each  other. 

EXAMPLES  TO  BE  WRITTEN. 


1.  Five  tenths. 

2.  Twenty-two  hundredths. 

3.  Ono  hundred  and  four  thou- 
sandths. 

4.  Two  units   and  one    hun- 
dredth. 

5.  Ono  thousand  six  hundred 
and  five  ten-thousandths. 

6.  Eighty-seven  hundred-thou- 
sandths. 

7.  Twenty-nine  and  a  half  ten- 
millionths. 

8.  Nineteen  million  and  one 
bill  ion  ths. 

9.  Seventy  thousand  and  forty- 
two  units  and  sixteen  hun- 
dredths. 

10.  Two  thousand  units  and 
fifty-six  and  a  third  inil- 
lionths 


11.  Four  hundred  and  twenty- 
one  tenths. 

12.  Six  thoueand  hundredths. 

13.  Eight  units  and  a  half  a 
hundredth: 

14.  Forty-eight  thousand  three 
hundred    and     five    thou- 
sandths. 

15.  Thirty-three     million    tec 
millionth*. 

16.  Four  hundred  thousandths 

17.  Four  hundred-thousandths. 

18.  One  unit  and  a  half  a  bil- 
lionth. 

19.  Sixty-six  thousand  and  thrw 
millionths. 

20.  Sixty-six  million  and  thref 
thousandths. 

21.  Thirty-four    and    a    Jhird 
tenths. 


REVIEW. — 141.  What  is  the  role  for  writing  decimals ?     Explain  it. 


102 


RAY'S   HIGHER   ARITHMETIC. 


22.  Forty-four  million  units  and 
four  millionths. 

23.  Two  hundred  and  eighteen 
thousand  and  six  billionths. 

24.  Xinety-six  thousands. 

25.  Ninety-six  hundreds. 
2G.    Ninety-six  tens. 

27.  Ninety-six  units. 

28.  Ninety-six  tenths. 

29.  Ninety-six  hundredths. 

30.  Ninety-six  thousandths. 


31.  Three   hundred  and    fifty- 
eight  thousand  and  six  ten 
millionths. 

32.  Two  million  millionths. 

33.  Four  million  units  and  four 
millionths. 

34.  Four  million  and  four  mil- 
lionths. 

35.  Fifty-thousand    and    seven 
hundred-thousandths. 

36.  Three   million  nnd  a  half 
billionths. 


ART.  142.  Decimal  Fractions  are  distinguished  from 
common  fractions,  by  not  having  written  denominators  ; 
and  from  whole  numbers,  by  the  decimal  point. 

The  denomination,  or  size  of  the  purls  in  any  decimal, 
depends  on  the  position  of  the  point;  so  that  the  set  of 
figures  which  expresses  but  one  value  as  a  whole  number, 
may,  as  a  decimal,  express  different  values,  according  to 
the  situation  of  the  point.  Great  care  must  be  exercised, 
then,  in  placing  the  point  correctly  and  distinctly. 

ART.  143.  PROPOSITION  I. — Decimal  ciphers  may  be  annexed 
to,  or  omitted  from,  the  right  of  any  number,  and  not  alter 

its  value. 


EXAMPLES. 

.25  =  . 250 
16  =  16. =16. 000 
19.08300  =  19.083 
200.00  =  200 


DEMONSTRATION. — The  ciphers 
themselves  are  of  no  value;  the 
other  figures  retain  their  places 
and,  consequently,  their  values. 
Hence,  the  value  of  the  number  is  not 
altered,  but  merely  its  denomination. 

REMARK. — Be  careful  that  the  ciphers  annexed  or  omitted  are 
decimal  ciphers. 

NOTE. — Annexing  or  omitting  ciphers  is  equivalent  to  multiplying 
or  dividing  both  terms  of  the  decimal  fraction  by  10,  100,  1000,  &c. 

ART.  144.  PROPOSITION  II. — If,  in  any  decimal,  the  point  be 
moved  to  the  RIGHT,  the  number  is  MULTIPLIED  continually  by  10 
as  often  as  a  figure  is  passed  over. 

RHVIBW. — 1-12.  How  arc  decimals  distinguished  from  common  frac- 
tions? How  from  whole  numbers?  How  is  the  size  of  the  parts  in  any 
decimal  known  ?  Why  should  groat  care  be  exercised  in  placing  the 
point  correctly  ?  143.  What  is  Proposition  1  ?  Frove  it.. 


REDUCTION    OF   DECIMALS.  1Q3 


DEMONSTRATION. — For  each  place  passed  over  EXAMPLES. 

by  the  point  in  moving  to  the  right,  every  figure  .0567 

is  advanced  one  step  in  the  scale  of  notation,  and  0.567 

is  worth  10  times  as  much  as  before;  hundreths  5.67 

are  changed  into  tenths,  tenths  into  units,  units  5  6.7 

into  tens,  tens  into  hundreds,  and  so  on;  hence,  567. 

the  proposition  is  true:  5670. 

ART  145.  PROPOSITION  III. — If,  in  any  decimal,  the  point 
be  moved  to  the  LEFT,  the  number  is  DIVIDED  continually  by  10 
as  often  as  a  figure  is  passed  over. 

DEMONSTRATION. — For  each  place  passed  EXAMPLES. 

over  by  the  point  in  moving  to  the  Ifft,  every  2340. 

figure  is  degraded  one  step  in  the  scale  of  nota-  234. 

tion,  and  is  worth  only  y1^  as  much  as  before;  2  3.4 

hundreds  are  changed  •into  tens,  tens  into  units,  2.34 

units  into  tenths,  tenths  into  hundredths,  and  so  .234 

on ;  hence,  the  proposition  is  true.  .0234 

NOTE. — If  the  point  is  to  be  moved  in  either  direction  over  more 
places  than  the  number  can  furnish,  supply  the  deficiency  by  taking 
in  ciphers,  as  in  the  last  examples  of  these  propositions. 

REMARK. — These  3  propositions  apply  to  any  whole  number, 
supposing  a  point  to  stand  on  the  right  of  its  units'  place. 


REDUCTION   OF    DECIMALS. 

ART.  146.  CASE  I. — To  convert  a  decimal  into  its 
simplest  equivalent  common  fraction. 

RULE. — Take  the  decimal  as  it  stands,  for  the  numerator, 
commencing  with  the  first  significant  figure;  for  the  denomi- 
nator, write  1  with  as  many  ciphers  annexed  as  there  are  decimal 
places;  then  reduce  this  fraction  to  its  lowest  terms. 

NOTE. — In  reducing  the  fraction  to  its  lowest  terms,  use  noi 
divisors  but  2  and  5;  since  they  are  the  only  prime  factors  of  10, 
and,  therefore,  the  only  prime  number  that  will  divide  the  denomi- 


REVIBW. — 144.  Stato  Proposition  2.  Prove  it,  145.  State  Proposi- 
tion 3.  Prove  it.  How  nro  deficient  places  supplied?  How  can  tbes« 
propositions  apply  to  whole  numbers?  146.  Wh«.t  is  mo  rule  for  reduo- 
\of  a  decimal  to  its  simplest  equivalent  common  fraction? 


104 


RAY'S   HIGHER   ARITHMETIC. 


nator,  which  is  either  10,  or  the  product  of  10's.  AVhen  the  nu- 
merator ends  in  any  odd  number  except  6,  neither  2  nor  6  wiJ 
divide  it,  (Art.  90),  and  the  fraction  will  be  in  its  lowest  terms. 

Reduce  .039375  to  its  equivalent  common  fraction. 

S o L u  T i  ON. — After  obtaining 
Ihe  common  fraction,  divide 
both  terms  by  5,  four  times  in 
luccesnion;  the  numerator  then 
ends  in  3,  and  the  fraction  can 
be  reduced  no  further.  (Note.) 

A  mixed  decimal  like  18.0067 
may  be  written  as  an  improper 


.039375=5) 


39375 
1000000 


200000 


'40000 
63 


'8000    1600 


Ant. 


fraction  V 
number  thus: 

By  this   rule,  a   complex   decimal  is   converted   into  a  complex 
common  fraction,  which  may  be  simplified  as  in  Art.  IM'J. 


8s1 


25 


5  1       . 

=  —  =  * —    Ans. 


Thus,  .084  =  ~L  = 

100        300        60         12 

The  rule  also  serves  to  change  a  part  of  a  pure  decimal  into 
a  common  fraction,  thereby  rendering  the  decimal  complex,  as, 
6.875=G.87j'f)==6.87A  ;  in  this  way  a  decimal  may  sometimes  be 
easily  reduced  to  a  common  fraction,  if  the  student  is  familiar 
with  the  aliquot  parts  of  100.  Thus, 


REDUCE   TO    COMMON    FRACTIONS. 


1. 

.25625   .    .    Ans.  TVo 

9. 

11.  OS     . 

A            111 

Ans.  ilyg 

2. 

.15234375     Ans.  ^  10. 

.390625     . 

Ans.  g4 

3. 

2.125     .     .    Ans.    21 

11. 

.1944^  .     . 

Ans.  Tf7c 

4. 

19.01750  Ans.  19?55 

12. 

.24$  .    .    . 

Ans.  \  I 

5. 

16.  00^    .     Am.  16500 

13 

.33}  .    .    . 

An*.  \ 

6. 

350.028$    Ans.  350^ 

14. 

.    Ans.  i 

7. 

.6666663    •     •  Ans.  f 

15 

.25    ... 

An*.    \ 

8. 

.003125       .    Ans.  rio 

16. 

.75   ... 

.     Ans.  1 

REVIEW. — 146.  What  divisors  only  need  be  used  in  reducing  to  the 
lowest  terms  ?  Why  ?  How  can  it  be  known,  by  inspection,  that  th« 
fraction  is  in  its  lowest  terms?  Why?  How  may  a  mixed  decimal  b« 
changed  to  a  common  fraction  ?  What  does  a  complex  fraction  bocoma 
by  the  application  of  the  rule?  How  can  a  pure  decimal  be  rendered 
complex  ?  Give  examples. 


REDUCTION    OF    DECIMALS. 


105 


17. 

.16|. 

.  Ans.  Js 

26. 

.911    . 

.  Ans. 

H 

18. 

.83K 

.    . 

.  Ans.  | 

27. 

.06i    . 

.  Ans. 

T'S 

19. 

.12-1  or 

.125 

Ans.  g 

28. 

.18!  . 

.  Ans. 

_3, 
1  8 

20. 

.37  A  or 

.375 

Ans.  | 

29. 

.314    . 

.     .  Ans. 

T68 

21. 

.624  or 

.625 

Ans.  | 

30. 

.43! 

.  Ans. 

7 

ra 

22. 

.87*jor 

.875 

Ans.  I 

31. 

.561    . 

4n*. 

r9a 

23. 

.081. 

,    4 

Ans.  YZ 

32. 

.68!    . 

.     .  Ans. 

11 

•24. 

.411  . 

.    . 

Ans.  fs 

33. 

.81*    . 

.  Ans. 

1! 

25. 

.581. 

. 

.An*.  75 

34. 

.93!    . 

.  Ans. 

is 

As  much  work  is  sometimes  saved,  by  using  a  common  fraction 
instead  of  its  equivalent  decimal,  the  pupil  should  be  familiar  with 
this  transformation. 

REMARK. — If  the  decimal  contain  a  great  many  places,  a  simple 
approximate  value  can  sometimes  be  obtained  by  using  the  first 
one  or  two  figures  only ;  for  example,  .3260873  is  nearly  y3^  =  | 
nearly;  and  1.7689=  ly^o  nearly  =  If  nearly. 

v 

CASE  II. — TO  CHANGE  ANY  FRACTION  INTO  A  DECIMAL. 

ART.  147.  If  the  fraction  has  10,  100,  1000,  &c.,  for 
its  denominator,  it  can  be  written  as  a  decimal,  by  writing 
the  numerator,  and  placing  the  point  so  that  the  number  of 
figures  on  the  right  of  it,  shall  be  equal  to  the  number  of 
ciphers  in  the  denominator. 

EXPRESS  IN  DECIMAL  FORM, 

JL  _JL  109  1 20056    52j    1600 
10'  100'  1000'  10000'  100000'  1000000' 
Ans.  .3,  .08,  .109,  12.0056,  .000521,  .001600, 

If  the  denominator  of  the  fraction,  is  not  10, 100, 1000, 
use  this 

RULE. — Divide  the  numerator  by  the  denominator,  annexing 
decimal  ciphers  to  the  former  as  they  are  needed ;  for  each  cipher 
annexed,  make  a  decimal  place  in  the  quotient. 

NOTES. — 1.  Before  annexing  ciphers  to  the  numerator,  be  careful 
to  place  a  point  on  its  right. 

RBVIKW. — 147.  How  can  a  fraction  whose  denominator  is  10,  100, 
1000,  Ac.,  be  written  as  a  decimal  ?  If  the  denominator  is  not  10,  100, 
1000,  Ac.?  What  should  be  done  before  annexing  the  ciphers? 


100  RAY'S    HIGHER   ARITHMETIC. 


2.  The  point  may  be  fixed  in  the  quotient  at  any  time,  by  making 
the  figure  last  written  in  the  quotient  occupy  the  same  decimal  place  as  tht 
figure  of  the  dividend  last  used;  in  doing   so,   prefix  ciphers   to  tht 
quotient,  if  necessary,  to  make  the  requisite  number  of  places. 

3.  The  operations  under  this  and  the  preceding  rule  serve  to  proY« 
each  other. 

Reduce  £  to  its  equivalent  decimal. 

or  ERA  n  ON, 

DEMONSTRATION. — The   numerator  with        o\^  nr>f\ 

o  )  /  .UU  U 
the  decimal  ciphers  annexed,  is  of  the  same 

value  as  before,  but  of  a  lower  denomination,  .875 

(Art.  143),  and,  when  it  is   divided   by  the       |  _-  .875  Ans 
denominator,  the  quotient  must  be  of  that  de- 
nomination also,  and  must  therefore  contain  the  same  number  of 
decimal  places. 

The  analysis  of  the  operation  is  as  follows :  g  =  g  of  7  units  =  ^ 
of  7000  thousandths  =  875  thousandths  =  .875 


REDUCE    TO  DECIMALS, 

1.  $     ....     =.75!    6.    f =.8 

2.  ^    ....   =.125      7.  29o9o      .     .     .    =.495 

3.  ^0  ....     =.051    8.  C5i  .     .     .  =  .078125 

4.  If.    .    .    =.46875 1    9.  ^35     .=.05078125 

5.  T59o<J    .    .  =  .005625|10.  7024  =  .0009765625 

The  rule  converts  a  mixed  number  into  a  mixed  decimal,  and  a 
complex  into  a  pure  decimal ;  thus,  9|=  9.375,  since  $  =  .375;  and 
.263^  =  .2612,  since  J?  =  a2 

11.  I6h    .    .    .   =16.5    13.    .015}     .     =.01525 

12.  42T3a  .     .=42.1875    14.    101.011=101.0175 

15.  75119g35 =75119.0375 

16.  2.00320- =2.00002125 

ART.  148.  Sometimes  annexing  ciphers  does  not  render 
the  numerator  exactly  divisible  by  the  denominator;  in 
that  case,  after  the  quotient  has  been  carried  out  as  far  as 
desirable,  the  sign  (+)  phts  is  annexed  to  show  that  there 
is  still  a  remainder.  Such,  having  no  end,  are  called 

REVIEW. — 147.  When  may  the  point  be  fixed  in  the  quotient?  How? 
Explain  the  example.  What  effect  does  the  rule  have  on  a  mixed  number  T 
On  a  complex  decimal  ? 


ADDITION    OF   DECIMALS.  107 


interminate  or  infinite  decimals;    thus,  2  =  .666666  +, 
an   interminate;  while   *  =  .125,  a   terminate  decimal. 

Sometimes,  when  the  remainder  omitted  in  an  inter- 
minate decimal  is  large  enough  to  give  the  next  quotient 
figure  more  than  5,  the  last  quotient  figure  is  written  1 
larger  than  it  really  is,  and  the  sign  ( — )  minus  is  annexed 
instead  of  (  +  )  plus,  to  show  that  the  quotient  is  written 
a  little  too  largo  ;  as,  §  =  .66+,  or  .67 — . 

1.  2\ =  .259259+ 

2.  10|f =  10.60417- 

3.  .065? =.0652857+ 

4.  430.18-jV =  430.1809+ 


ADDITION   OF   DECIMALS. 

ART.  149.  RULE. —  Write  the  numbers  to  be  added  so  that 
figures  of  the  same  denomination  may  be  in  columns;  add  as  in 
wJwk  numbers,  commencing  at  the  right,  and  point  off  the  result 
to  agree  with  that  one  of  the  given  numbers  having  the  most 
decimal  places. 

PROOF. — Same  as  in  addition  of  whole  numbers. 

NOTE. — Complex  decimals,  if  there  are  any,  must  be  made  pure, 
(Art.  147),  as  far,  at  least,  as  the  decimal  places  extend  in  the  other 
numbers.  If,  after  (his,  there  are  common  fractions  in  the  right- 
hand  column,  add  them;  or,  neglect  them,  using  the  signs  -f-  or  —  as 
in  Art.  148. 

Add  23.8  and  17^  and  .0256  and  .41| 

SOLUTION.— After    reducing,   set  OPERATION. 

figures  of  the  same  order  in  column  ;  0  Q  8 

then  add  and  carry  as  in  whole  uum-  -•  HI  1  IT'K 

bers.     As  the  right-hand  figures,  when  ft  9  ^  fi 

added,    must    make    the    right-hand  A-\~_        *4irP'- 
figure  of  the  answer  of  the  same  de- 
nomination as  those  in  the  column,  Ans.  41.74223 
the  point  should  be  fixed  as  the  rule  directs. 


REVIEW. — 148.  When  tho  quotient  can  not  be  made  exact,  what  is 
done  ?  What  arc  such  decimals  called  t  Why  ?  When  may  tho  sign 
minus  bo  usod  1 


108  RAY'S   HIGHER    A  UtTHMETIC. 


1.  Find  the  sum -of  1  +  .9475  An*.  1.9475 

2.  Of  1.33  J  added  to  itself  twice.  Ana.  4. 

3.  Of  14. 034,  25,    .000062^,    .0034 

Am.  39.0374625 

4.  Of  83    thousandths,  2101  hundredths,  25  tenth* 
nd  94s  units.  AM.  118.093 

5.  Of  .16i,  .37*,  5,  3.41,  .0001     Am.  8.980^ 

6.  Of  4  units,  4  tenths,  4  huudredths.       Ans.  4.44 

7.  Of  .1H  +.66661  +  .2222225  Ami. 

8.  Of  .141,  .0181,  920,  .0139!      Am.  920.1754 

9.  Of  16. 008|,  .00741,    .2§,    .00019042i 

Am.  16.299768199? 

10.  Of  .675,  2  millionths,  64g,  and  3.49000107 

Am.  68.29000307 

11.  Of  four  times  4.0671  and  .000}      ,l«s.  16.272 

12.  Of  216. 86301,  48.1057,  .029,  1.3,  1000. 

Ans.  1266.29771 

13.  Add  35  units,  35  tenths,  35  hundredths,  35  thou- 
sandths. Ans.  38.885 

14.  Add  ten  thousand  and  one  millionths;    four  hun- 
dred-thousandths;   96    hundredths;    forty-seven    million 
sixty  thousand  and  eight  billionths.  Am.  1.017101008 


SUBTKACTION  OF   DECIMALS. 

ART.  150.  RULE — Write  the  subtrahend  under  the  minuend, 
placing  figures  of  the  same  denomination  in  columns,  ffubtract 
as  in  whole  numbers,  commencing  at  the  right,  and  point  the 
result  as  in  addition  of  decimals. 

PROOF. — As  in  subtraction  of  whole  numbers. 

NOTES. — If  either  or  both  of  the  given  decimals  be  complex, 
proceed  as  directed  in  the  note  to  last  rule. 


RRYIEW. — 149.  What  is  the  rule  for  adding  decimals?  The  proof? 
What  must  be  done  with  complex  decimals  ?  If  there  are  common  frac- 
tions in  .the  right-hand  column,  what  should  bo  done  ?  Explain  the  ex- 
ample.  150.  What  is  the  rule  for  subtraction  of  decimals?  The  proof? 


SUBTRACTION    OF   DECIMALS. 


2.  If  the  minuend  has  not  as  many  decimal  places  as  the  subtra- 
hend, annex  decimal  ciphers  to  it,  or  suppose  them  to  be  annexed 
until  the  deficiency  is  supplied. 

From  68  subtract  2.05*7 

SOLUTION.  —  Write  the  numbers  as  the  rule  /•   o 

directs;  suppose  ciphers  to  be  annexed  to  the  8,  o'ft  ^^7 

and  subtract  as  in  whole  numbers;  saying  7  from 
10  leaves  3,  carry  1 ;  6  from  10  leaves  4,  and  so  on.      Ans.  4.743 

From  13.2563  subtract  6.77i 

-|    9    f)  c  ft  6 

In  this  example,  the  complex  decimals  o .  ^  o 

are  rendei'ed  pure  to  the  same  extent,  and      6.77s  =  6.773s 
the  common  fractions  subtracted.  ^        g  483^1 

EXAMPLES   FOR   PRACTICE. 

1.  Subtract  8.00717  from  19.54    Ans.  11.53283 

2.  3  thousandths  from  3000.  Ans.  2999.997 

3.  72.0001  from  72.01  Ans.  .0099 

4.  Subtract ..93 A  from  1.1691  Ans.  .238^ 

5.  How  much  is  19  less  8.999i?       Ana.  lO.OOOf 

6.  How  much  less  is  .041  than   .4?.        Ans.  .35f 

7.  How  much  is  .65007  —  2?  Am.  .15007 

8.  What  is  2| — If  in  decimals?  Ans.  .95 

9.  Take  .007601  twice  from  .02      Ans.  .004798 

10.  Take  1.98i  three  times  from  6.  Ans.  .05 

11.  Subtract  1  from  1.684  Ans.  .684 

12.  |  of  a  millionth  from  .000$       Ans.  .000443JS 

13.  H  hundrcdths  from  49!  tenths.       Ans.  4.9225 

14.  10000  thousandths  from  10  units.  Ans.  0 
15  24^  tenths  from  3701  thousandths.  Ans.  1.251 

16.  Is  units  from  1875  thousandths.  -4ns.  0 

17.  TJ  of  a  hundreth  from  yg  of  a  tenth.          Ans.  0 

18.  64g  hundredths  from  100  units.      Ans.  99.35§ 


UK  VIEW. — 150.  What  must  be  done  with  complex  decimals?  If  the 
minuend  do  not  contain  as  many  decimal  places  as  the  subtrahend,  what 
must  be  done?  Explain  tho  examples. 


110  RAY'S   HIGHER   ARITHMETIC. 


MULTIPLICATION   OF   DECIMALS. 
ART.  151.    RULE. — Multiply  as  in  whole  numbers,  and  point 
tne  product,  so  that  it  shall  liave  as  many  decimal  places  as  the 
multiplicand  and  multiplier  together. 

PROOF.  —  As  in  multiplication  of  whole  numbers. 
HEM  ARK.  — If  the  product,  has  not  as  many  decimal  places  u 
c-iuired,  supply  the  deficiency  by  prefixing  ciphers. 

Multiply  2.56  by  .184 

DEMONSTRATION. — Express     256        184          47104 
tne  decimals  as  common   frac-      1  ft  ft       1  ft  ()7)  ==  1  00000 
tions ;  their  product  will  be  the 
product  of  their  numerators  di-  Hence, 

vidcd  by  the  product  of    their      2.56X  .184  =  . 47104 
denominators,  (Art.  130).   Writn 

the  fractions  in  decimal  form;  the  denominator  of  the  product  has 
as  many  ciphers  as  both  the  other  denominators:  and  as  each  of 
these  ciphers  make  a  decimal  place,  (Art.  136),  the  product  will  hav« 
as  many  decimal  places  as  both  factors. 

EXAMPLES    FOR   PRACTICE. 

1.  1  X  .1          =.1 

2.  16  X  .031 =  .531 

3.  .01  X  .U       =.0015 

4.  .080X  80 =6.4 

5.  37. 5  X  821 =3093.75 

6.  64.01  X  .32 =20.4832 

7.  48000.  X  73 =3504000. 

8.  64. 66;  X  18 =1164. 

9.  .56',  X  .OSi'g =   .0172J-! 

10.  738  x  120.4 88S&5.2 

11.  .0001X1.006       =.0001006 

12.  34  units  X. 193 =6.562 

13.  27  tenths  X.4i =1.134 

14.  43. 7004  X. 008 =.3496032 

15.  21.0375  X  4.441  .  =93.5 


REVIEW. — 151.  What  is  the  rule  for  Multiplication  of  Decimals?  Thi 
proof?  How  are  deficient  places  in  the  product  to  bo  supplied?  Prort 
the  rule. 


5.  V  9300. 


MULTIPLICATION    OF   DECIMALS.  1 1 


16.V  9300. 701  X  251    ....     =2334475.951 

17.  430.0126X4000 =1720050.4 

18.  .059  x  .059  X  .059  .    .    .    .  =  .0002u5379 

19.  42  units  X  42  tenths =  176.4 

20.  21  hundredths  X  600 =13.5 

21.  7100  X  s  of  a  millionth.      .     .       =  .0008876 

22.  26  millions  X  26  millionths =67C. 

23.  2700  hundredths  X  60  tenths  .     .     .     .  =  162 

24.  What  denomination    will   a  figure   in    the   tenths 
place,   multiplied    by   a    figure  in  the  hundredths'  place, 
eive?  Ans.  .1  X  .01  =  .001  or  thousandths. 

25.  A  figure  in  the  units'  place,  by  one  in  the  hun- 
dred-thousandths' place?        Ans.  Hundred-thousandths. 

26.  1  thousandth  by  1  thousandth?  Ans.  Millionths. 

27.  1  hundred  by  1  tenth?  Ans.  Tens. 

28.  1  in  thousands'  by  1  in  tenths'?  Ans.  Hundreds. 

CONTRACTED  MULTIPLICATION   OF    DECIMALS. 

ART.  152.  The  methods  of  contracting  multiplication 
in  whole  numbers,  apply  also  to  multiplication  of  decimals. 
It  is  only  necessary  to  call  attention  to  two  cases. 

CASE  I. — TO   MULTIPLY   A   DECIMAL   BY  10,    100,  1000,   &C. 

RULE. — Remove  the  decimal  point  of  the  multiplicand  to  thv 
right,  over  as  many  places  as  there  are  cipJiers  in  the  multiplier; 
the  result  will  be  the  product  required. 

NOTES. — 1.  If  there  are  not  as  many  places  to  the  right  of  the 
point  as  are  required  in  the  operation,  supply  the  deficiency  by 
taking  in  ciphers,  unless  the  multiplicand  be  a  complex  decimal; 
in  that  case,  the  proper  figures  should  be  ascertained  and  set 
down,  (Art.  147). 

2.  The  rule  given  in  Art.  64,  for  multiplying  a  whole  number  by 
10,  100,  1000,  &c.,  is  a  particular  case  of  this,  since  a  whole  num- 
ber may  be  regarded  as  a  decimal,  the  point  being  on  the  right 
of  the  units'  place,  (Art.  137). 


REVIEW. — 152.  What  methods  of  Contracted  Multiplication  euro  used 
for  decimals?  What  is  the  1st  Case?  Give  the  rule.  How  are  deficient 
places  supplied  ?  How  does  this  rule  apply  to  whole  cumbers  ? 


112  RAY'S   HIGHER   ARITHMETIC. 


. — To  multiply  by  10,  100,  1000,  &c.,  is  th« 
same  as  to  multiply  continually  by  10;  this  is  done  by  moving  the 
point  to  the  right,  over  as  many  places  as  there  are  ciphers  in 
the  multiplier,  (Art.  144). 

EXAMPLES   FOR   PRACTICE. 

1.  56.x  100 =5600. 

2.  .075  X  100     .    .    .    .    • =7.5 

3.  .01!  X  1000 =17.5 

4.  16. 083  X  10 =160.83 

5.  10.341x100000 =1034400. 

6.  98.0471  X 1000000     .    .    .    .=980473331 

ART.  153.  Whenever  the  product  of  two  decimals  is  not 
required  to  contain  figures  below  a  certain  order,  the  work 
may  be  shortened. 

CASE    n.  —  TO    MULTIPLY,    RESERVING   A   CERTAIN    NUMBER 
OP   DECIMAL   PLACES    IN   THE    PRODUCT. 

RULE. —  Count  off  in  the  multiplicand,  the  number  of  decimal 
places  to  be  reserved,  draw  a,  vertical  line  through  the  lowest,  and 
write  the  multiplier  so  that  its  units'  figure  shall  fall  on  this  line. 
Begin  at  tlie  left  of  the  multiplier  to  form  the  partial  products, 
always  starting  at  that  figure  of  the  multiplicand  which  is  as 
far  on  one  side  of  the  line,  as  the  figure  of  the  multiplier  then 
in  use  is  on  the  other  side,  carrying  the  tens,  however,  obtained 
by  multiplying  the  next  lower  figure. 

Set  the  right-hand  figures  of  these  partial  products  in  a 
column,  add  and  point  off  the  number  of  decimal  places  required. 

NOTES. — 1.  The  multiplicand  should  extend  one  figure  further 
to  the  right  of  the  line  than  the  multiplier  does  to  the  left  of  it. 
To  accomplish  this,  it  may  be  necessary  to  annex  decimal  ciphers, 
or  to  convert  a  common  fraction  into  a  decimal. 

2.  If  the  multiplier  contain  a  common  fraction,  and  it  becomes 
necessary  to  multiply  by  it,  start  at  the  same  figure  of  the  mul- 
tiplicand as  in  the  previous  multiplication. 

REMARK. — In  carrying  tens  from  the  nearest  rejected  figure  of 
the  multiplicand,  carry  1  ten  also  for  any  number  of  units  over  6; 
thus,  for  65,  carry  6;  for  18,  carry  2. 

REVIEW. — 152.  Demonstrate  it     153.   What  is  Cafe  2?     The 
How  far  should  the  multiplicand  extend  to  the  right  'if  tho  line? 


MULTIPLICATION  OF   DECIMALS.  113 

Multiply  1.89361  by  3.5672,  reserving  3  decimals  in 
the  product. 

S  c  L  u  T  i  o  jr.— Count  off        8nORT  *MnOD-         OBDINARY  METHOD. 
3  decimals  in  the  multi-        1.89361  1.89361 

plicand,  draw  a  vertical      3.5  672  3.5  672 

line  through  the  lowest 


(3\  and  proceed  thus: 
Commencing   with   3, 
iLe   left-hand  figure    of 
the  multiplier,  and  3  in          ^      , 
the  multiplicand,  which  ' 


13 
113 

946 
5680 


6.754 


378722 

25527 

6166 

805 

83 


885592 


are  both  on  the  line,  say, 
3  times  3  are  9 ;  but  3  times  6  (the  next  lower  figure)  are  18,  which 
is  nearer  2  tens  than  1  ten;  carrying  these  2  tens  to  9  we  have  11, 
set  down  1  and  carry  1.  Keeping  to  the  left,  the  first  partial  pro- 
duct is  6681.  Next,  take  5  in  the  multiplier,  and  start  at  9  in  the 
multiplicand,  carrying  2  for  the  15  obtained  by  multiplying  the 
next  lower  figure,  3. 

The  second  partial  product  is  947,  whose  first  figure  is  set  under 
the  first  figure  (1)  of  the  previous  product.  Thus  go  on,  using  6 
and  7  in  the  multiplier  successively,  and  starting  at  8  and  1  in 
the  multiplicand.  As  the  figures  of  the  multiplicand  toward  the 
left  are  then  exhausted,  the  operation,  after  adding  and  pointing 
off  3  figures,  is  complete. 

DEMONSTRATION. — The  reason  of  the  rule  consists  in  the  fact, 
that  all  the  multiplications  which  would  give  rise  to  denominations 
lower  than  those  required  in  the  product,  are  omitted. 

The  rule  will  not  always  give  the  lav.)  figure  correct. 
To  insure  perfect  accuracy,  reserve  one  more  figure  than  « 
required,  and  omit  the  last  figure  of  the  product. 

What  is  the  product  of  .0543G3TT  by  6458.19  true 
to  3  decimal  places? 

SOLUTION. — Reserve  4  decimals,  one  more  than  is  required; 
tarry  out  the  figures  of  the  multiplicand,  (Note  1) :  point  off  4 
places  in  the  product,  351.0906,  writing  the  next  to  the  last,  1,  in- 
stead of  0,  because  the  figure  omitted  is  over  6.  This  gives  351.091 
for  the  answer. 


REVIEW. — 153.  How  a-e  deficient  places  supplied?  What,  if  tie  mul- 
tiplier contains  a  common  traction?  In  carrying  tens  from  the  rejected 
figures,  what  directions  must  be  observed?  Solve  the  example.  What 
denomination  in  the  product,  is  obtained  by  multiplying  the  marked  figure 
df  the  multiplicand  by  the  units'  figure  of  the  multiplier  ?  Ant.  The  lowest 
10 


114 


RAY'S  HIGHER   ARITHMETIC. 


1. 

2, 

EXAijfPDBS/                      Decimal*  reserved.          ANSWERS 

4.750}  X  .002.8.6^6.  fuitU5                  .01361 
804.571  X  17.081^2          4         13743.2315 

3. 

75.062x460.8917 

2 

34595.45 

NOTE.—  Take  75.062  for  the 

multiplier. 

4. 

9.012x48.75 

1 

439.3 

5. 

4.804136  x  .010759 

6 

.051688 

6. 

814A33  x  26f  f 

3 

21813.475 

7. 

702.  61  x  1.258378 

3 

884.020 

8. 

849.93!  x  .0424444 

3 

36.075 

9. 

880.  695  x  131.  72  true 

to,  units. 

116005. 

10. 

.025381  X  .004907 

5 

.00012 

11. 

64.01082  X  .03537' 

6 

2.264063 

12. 
13. 

7.24691  x  $1.4632 
.  681472  x  .01286 

3 
5 

590.324 

.00876 

4. 

7.  9443  x  3.69 

4 

29.3150 

15. 

.053497  x  .047126 

6 

.002521 

16. 

1380.  37^X  .2341 

2 

324.16 

DIVISION    OF    DECIMALS. 

ART.  154.  RULE. — Make  the  decimal  places  of  the  dividend 
as  many  as  those  of  the  divisor,  if  they  are  less.  Divide  as  in 
whole  numbers,  annexing  other  decimal  figures  to  the  dividend  as 
they  are  needed;  point  the  quotient  so  that  it  shall  have  as  many 
decimal  places  as  the  dividend  has  MORE  than  tlie  divisor. 

PROOF. — Same  as  in  Division  of  whole  numbers. 

NOTES. — 1.  To  extend  the  dividend,  decimal  ciphers  i\ro  used; 
but,  if  the  dividend  is  a  complex  decimal,  make  it  pure. 

2.  If  the  quotient   has  not  the  number  of  figures  required  for 
decimal  places,  prefix  ciphers. 

3.  If  the  dividend  has  the  same  number  of  decimal  places  as  the 
divisor,  the  quotient  is  units. 


R  E  vi  E  w. — 153.  What,  by  multiplying  each  figure  of  tho  multiplier  by 
the  figure  of  tho  multiplicand  as  far  on  the  other  side  of  the  lino  ?  Ana.  The 
earne.  What  multiplications  are  omitted?  154.  What  is  tho  rulo  for  di- 
vision of  decimals?  The  proof?  How  are  deficient  places  in  the  dividend 
supplied?  How  in  the  quotient  T 


DIVISION  OF   DECIMALS.  115 

4.  Make  complex  decimals  pure;  or,  divide  them  like  common 
mixed  numbers ;  or,  multiply  both  by  the  least  common  multiple 
of  the  denominators  of  the  common  fractions,  and  then  divide. 

6.  If  the  division  is  not  exact,  it  may  be  continued  by  annexing 
other  decimal  figures  to  the  dividend;  or  the'  remainder  may  be 
written  with  the  divisor  under  it,  as  a  common  fraction. 

Divide  .50312  by  .19 

DEMONSTRATION.— Since  the  di-      .19). 50     12(2.648 
Tidend  is  the  product  of  tho  divisor  1 Z  O  Ans. 

and  quotient,  it  must  have  as  many 
decimal  places  as  both  of  them,  (Art  152 

161);  hence,  the  quotient  must  have 

decimal  places  enough  to  make  with  those  of  the  divisor  as  many  aa 
are  in  the  dividend,  which  is  just  as  many  as  the  dividend  has  more 
than  the  divisor. 

Divide  24  by  3.2 

Annex  a  decimal  cipher  to  the  dividend,      3.2)24.00(7.5 
to  make  it  have  as  many  decimal  places  as  1.60 

the  divisor;  afterward,  annex  another,  to  0 

continue  the  division. 

Divide  .07  by  21.6 

In  long  division,  annex  the  21.6)  -OTOOf. 0032407+ 
ciphera  to  tho  remainders  as  QQA 

they   occur,    reckoning    them  i  /»  A  A 

still  as  decimal  places  of  the 
dividend;   here   the   dividend 

has  eight  decimal  places,  counting  the  ciphers  annexed  to  the  re- 
mainders, 

1.  Divide  .002J8  by  .061  Ans.  .0375 
SUGGESTION. — Convert  the   numbers  into  the  pure  decimals 

.002475  and  .066 ;  or,  multiply  both  by  40,  the  least  common  mul- 
tiple of  the  denominators  5  and  40,  making  the  dividend  .099,  and 
the  divisor  2.64:  use  the  latter  method  generally. 

EXAMPLES  FOR  PRACTICE. 

2.  3-J-.18I =16. 

3.  4. 2-=-. 311 =13.44 


REVIEW. — 154.  If  the  dividend  and  divisor  have  the  earn*  number  of 
decimal  places,  what  is  the  quotient?  How  may  complex  decimals  be 
divided  ?  If  the  division  is  not  exact,  what  may  be  done  ?  Demonstrate  th« 
rnlo.  Explain  the  examples. 


116  RAY'S   HIGHER   ARITHMETIC. 


4.  63-MOOO          =.01575 

5.  3.15-f-375 =  .0084 

6.  1.008-^-18        =.056 

7.  4096-f-.032 =128000. 

8.  9.7-T-97000 =.0001 

9.  .9^-. 00075       =1200. 

10.  13-^-78.12^       =.1664 

11.  12.9-j-8.256          =1.5625 

12.  81.2096-r-l.28 =63.445 

13.  12755^-81632 =.15625 

14.  2401H-21.4375 =112. 

15.  21.13212-r-.916        =23.07 

16.  36. 72672-j-. 5025     .....     =73.088 

17.  2483.25-^-5.1562$  V*    .    .    .    .=481.6 

18.  142. 0281-T-9. 2376 =15.375 

19.  l-r-100  =  .01 

20.  lO.lH-17          =  .59412- 

21.  VOOl-r-100        =.00001 

22.\   08|-r-.12i       =.66f  =  | 

23.  V.  0001 -5-.  01 =.01 

24.  95.3-r-.264 =360.984848+ 

25.  1000 -:-. 001 =1000000. 

26.  Ten -^-1  tenth         =100. 

27.  . 000001 -f-. 01 =.0001 

28.  .OOOOl-f-1000 =.00000001 

29.  16. 275  -=-.41664      =39.0625 

30.  1  ten-millionth  -j-1  hundreth    .          =.00001 

v 

ART.  155.  If  the  dividend  is  less  than  the  divisor, 
the  quotient  may  be  expressed  as  a  common  fraction,  by 
taking  the  dividend  for  the  numerator,  the  divisor  for  the 
denominator,  making  both  terms  have  the  same  number 
of  decimal  places,  and  then  omitting  the  points,  which  is 
equivalent  to  multiplying  them  by  the  same  number, 
(Art.  144);  thus,  3.737  divided  by  99.9,  equals  ^3?7 

_     3-737 3737 101 

—  ?  0-000— -D9900  — 270  0- 

REVIEW. — 155.  If  the  dividend  be  less  than  the  divisor,  how  may  th« 
quotient  be  written? 


DIVISION  OF   DECIMALS.  H7 


1. 

2. 

.75- 
1-f- 

5 

-2. 

.5 

125 

=  i  f 

_    2 
-   1  1 

3. 
4. 

.181-5-2. 
.0411-^ 

4 
.151 

"* 

=  75 

AET.  158.  To  find  the  denomination  of  the  1st  quo- 
tient figure  when  obtained,  count  off  in  the  dividend,  at 
many  decimal  placet  as  are  in  the  divisor,  placing  a  dot 
after  the  last ;  numerate  from  this  dot  as  a  decimal  point, 
either  way,  up  to  that  figure  of  the  dividend,  under  which 
the  right-hand  figure  of  the  ~ist  product  falls.  This  will 
give  the  denomination  of  the  1st  quotient  figure. 

Thus,  in  dividing  .07  by  21.6,  (page  115),  mark  the  dividend, 
.0*700;  numerate  from  the  dot  as  a  decimal  point  to  the  right  as 
far  as  the  0,  under  which  the  right-hand  figure  of  the  first  product 
falls:  this  gives  thousandths,  which  is  the  denomination  of  the  1st 
quotient  figure,  3. 

CONTRACTED   DIVISION  OF  DECIMALS. 

ART.  157.  The  methods  of  contracting  division  in 
whole  numbers  apply  also  to  decimals.  Only  three  cases 
need  be  noticed. 

CASE  I. — TO  DIVIDE  A  DECIMAL  BY  10,   100,   1000,  &c. 

RULE. — Remove  the  decimal  point  of  the  dividend  to  the  left, 
over  as  many  places  as  there  are  ciphers  in  the  divisor;  the 
result  will  be  the  required  quotient. 

NOTE. — If  there  are  not  as  many  places  to  the  left  as  are  re- 
quired, prefix  ciphers. 

REHAB K. — The  rule  in  Art  67,  for  dividing  a  whole  number  by 
10,  100,  1000,  &c.,  is  a  particular  case  of  this,  since  a  whole  num- 
ber may  be  considered  a  decimal  with  a  point  on  its  right.  (Art.  137.) 

Divide  68.075  by  10000.  Ant.  .0068075 

DEMONSTRATION. — To  divide  by  10,  100,  1000,  &c.,  is  the  same 
as  to  divide  continually  by  10.  This  is  done  by  moving  the  point 
to  the  left,  over  as  many  places  as  there  are  ciphers  in  the  divisor, 
(Art.  145.) 

R  p.  v  i  K  w. — 155.  Before  reducing  the  common  fraction,  what  must  be 
done?  Why?  156.  How  is  the  order  of  any  quotient  figure  determined 
ai  soon  aa  it  is  set  down?  157.  What  methods  of  contraction  are  used 
in  division  of  decimals?  What  is  the  1st  case?  The  rule?  How  ar« 
deficient  places  supplied?  Why  can  the  rule  be  applied  to  whole  num- 
bers ?  Prove  the  rule. 


118 


RAY'S   HIGHER   ARITHMETIC. 


1. 

2. 

3. 
4. 
5. 
fi 

65  —  1000          .     .     .     , 

=.065 

.072g  —  10        .     .    .    , 

,    .    .    .    .    =.00729 

i-^ioooooo      .  .  , 

m  e  t    _  .000001 

200ll2-^-100        .    . 

—20.012 

93000-^-1000        .    . 
4.472-^10000 

.    .   =93.000  =  93. 
.    -.0004*772 

ART.  158.  It  often  happens  that  the  quotient  is  not 
required  to  contain  decimal  figures  below  a  certain  de- 
nomination; if  so,  the  work  may  be  shortened. 

i  4.SE    II.  —  TO   DIVIDE,    RESERVING   A   CERTAIN    NUMBER   OP 
DECIMALS   IN   THE    QUOTIENT. 

RULE.  —  Mark  that  figure  of  the  dividend,  whose  denomination 
would  result  from  multiplying  a  unit  of  the  highest  denomina- 
tion in  the  divisor,  by  a  unit  of  the  lowest  denomination  required 
in  the  quotient, 

Divide  as  usual,  until  tJiis  figure  is  reached  ;  then,  stop  bring- 
ing down  from  the  dividend,  and  at  each  subsequent  division, 
drop  a  figure  from  the  divisor,  carrying  for  its  tens. 

Continue  thus,  until  the  divisor  is  reduced  to  a  single  figure, 
and  then  point  off  the  quotient  as  required. 

NOTE.  —  If  the  marked  figure  of  the  dividend  is  reached  at  the  first 
multiplication,  mark  the  figure  of  the  divisor,  whose  product  by  the 
first  quotient  figure  falls  under  the  marked  figure  of  the  dividend; 
and  at  the  next  step,  reject  this  figure  with  those  on  its  right. 

REMARKS.  —  n.  If  the  dividend  has  no  figure  of  the  denomina- 
tion to  be  marked,  annex  ciphers  to  it,  or  continue  it  further,  if  it 
is  an  interminate  decimal,  until  it  does. 

2.  In  carrying  tens  from  the  rejected  figures,  observe  the  direc- 
tions in  Case  2  of  contracted  multiplication  of  decimals.  (Art. 
153,  Rem.) 

Divide  1.078543  by  319.562  true  to  5  decimal  figures. 


SHORT  METHOD. 


319 


073  542(.00337 
958  6'86 


120 
96 

24 
_22 

2 


ORDINARY    METHOD. 

319.562)1.078j543(.00337 
958!686 

11918570 

95^8686 

2398840 
22  36934 

1161906 


DIVISION    OF   DECIMALS. 


SOLUTION.  —  The  highest  denomination  in  the  divisor  is  hundreds, 
the  lowest  required  in  the  quotient  is\hundred-thousandths  ;  and 
100  X  -00001  =.001;  therefore,  mark  the  thousandths'  figure  (8)  of 
the  dividend.  As  this  marked  figure  is  reached  at  the  first  mul- 
tiplication mark  9  in  the  multiplier,  since  it  gives  the  figure  that 
falls  under  8  in  the  dividend. 

Cross  the  rejected  figures  of  dividend  and  divisor;  cross  on* 
more  from  the  divisor  at  every  new  multiplication,  carrying  tent 
as  directed:  837,  with  the  necessary  ciphers  and  point  prefixed, 
is  the  quotient  required. 

DEMONSTRATION.  —  The  reason  of  the  rule  consists  in  the  fact, 
that  all  operations,  which  would  involve  denominations  lower 
than  those  required  in  the  quotient,  are  omitted. 

EXAMPLES.  Decimals  reserved.     ANSWERS. 

1.  lH-2.6j[8^  3  .373 

2.  10.  00371  V.  056248  2  177.85 

3.  .187564  -=-.00043129  true  to  units.        435. 

4.  .  007516362  -r-  652.  18  8  .00001152 

5.  1000.86-7-3.1415926  7  318.5836381 

6.  61.0598314-^4278  6  .014273 

7.  421.  33i-=-  9.104f  4  46.2778 

8.  100  -f-  3.  7320508  5  26.79492 

9.  7912.5043-^-181.34  1  43.6 

10.  .9l3-f-216.52  .10    .0042233717 

11.  555  -T-  123456789  10    .0000044955 

12.  1-v-  111111  6  .000009 

CASE   III.  —  TO   DIVItflTBY   A   DECIMAL    LITTLE    LESS    THAN 
1,  RESERVING   DECIMALS  IN  THE   QUOTIENT. 

ART.  159.  RULE.  —  Multiply  the  dividend  by  what  the  divisor 
wants  of  being  a  unit;  multiply  this  product  in  like  manner, 
and  continue  so  until  the  product  becomes  too  small  to  affect  ihs 
result  as  required;  then  add  to  obtain  the  quotient. 

Divide  3815.64  by  .994,  reserving  2  decimals  in  the 
quotient. 


REVIEW. — 158.  What  IB  Case  2?  Tho  rule?  Why  should  no  d»- 
Dominations  of  tho  dividend  be  used  below  the  ono  marked?  Ans.  Be- 
cause, when  divided  by  the  divisor,  they  give  lotfer  denominations  than 
are  required  in  tho  quotient 


HAY'S   HIGHER   ARITHMETIC. 


SOLUTION— Multiply    3815.64   by   .000,    the      3815.64 
difference  between  .994  and  a  unit;  write  the  22.894 

product,  22.894,  neglecting  all  denominations  .137 

below  thousandths.     Do  the  same  to  this  pro-  1 

duct,  and  to  the  next.     The  rest  of   the  pro-       q  Q  q  o  a~n ro 
ducts  are  too  small  to  affect  the  answer  as  re-  o .  0  <  ^ 

quired;   therefore,  add  and  obtain  the  quotient, 
8838. (57,  true  to  2  decimals. 

DEMONSTRATION. — The  operation  and  demonstration  are  similar 
to  those  in  Art.  G9,  for  the  corresponding  case  of  whole  numbers, 
except  that  no  remainders  occur,  the  product  being  extended  in 
decimals  as  far  as  is  necessary  to  obtain  the  correct  answer. 

EXAMPLES.  Decimals  reserved.  ANSWERS. 

1.  1000-:- .98  2  1020.41 

2.  6215. 75-j-.  99£  3  6246.985 

3.  28012 -f-. 993  2  28209.47 

4.  52546. 35  ^-.99!  3  52678.045 

5.  4840 -=-.9875  2  4901.27 


X.   CIRCULATING   DECIMALS. 

ART.  160.  Many  common  fractions,  when  transformed, 
become  interminate  decimals.  (Art.  148.)  These  have 
some  curious  and  useful  properties  worth  considering. 

PROPOSITION   I. 

The  only  common  fractions  which  can  be  changed  into 
terminate  decimals,  are  those  which,  reduced  to  their  lowest 
terms,  have  no  factors  but  2  and  5  in  their  denominators. 

DEMONSTRATION.  — The  numerator  must  EX  AMPLES. 

contain  all  the  prime  factors  of  the  denomi-      _3._ —  .QQ9375 
nator   to  be  divisible  by  it.     Every  cipher 
annexed  to  the  numerator  multiplies  it  by        5    =-3125 
10,  introducing  2  and  5,  the  factors  of  10, 

as  factors  of  the  numerator.    If  the  denomi-      _A.  =  .666664- 
nator  has  no  factors  but  2's  and  6's,  enough 

ciphers  may  be  annexed  to  the  numerator  to  =.72727-f- 

give  it  as  many  2's  and  5's  for  factors  as  the 

denominator,  and  then  the  quotient  is  exact.  But  if  the  denomi- 
nator have  any  factor  besides  2  and  6,  this  factor  never  can  be  intro- 
duced into  the  numerator  by  annexing  ciphers,  for  2  aud  5  are  the 
only  factors  that  can  be  so  introduced.  In  such  cases,  the  exact 
division  is  not.  possible. 


CIRCULATING    DECIMALS.  12] 

Tell  whether  the  following  common  fractions  can  b« 
changed  into  terminate  or  interminate  decimals:; 

JL         JLJLil.?!L10§21li.6§l 
9*        8'       82'       12'       162'       112'      30'      288;      87  J 

PROPOSITION  II. 

ART.  161.  Every  interminate  decimal  arising  from  lh« 
transformation  of  a  common  fraction  will  le  found,  if  the 
division  5e  carried  far  enough,  to  contain  the  same  figure,  or 
%et  of  figures,  repeated  in  the  same  order  without  end. 

DEMOXSTKATION. — In  converting 

y    into  a  decimal,  the    remainders  EXAMPLES. 

»re  successively  3,  2,  6,  4,  6,  1,  and      y=.  142857142857+ 
as  these  are  all  the  whole  numbers 
less  than  7,  the  next  remainder  must      J*_=  5555-1- 
bo  one  of  these  repeated;  on  trial  it       9 
is  found  to  be  8,  to  -which  if  a  0  bo       2        n7,lfi7A 
added,  we  have  the  same  dividend      27=  •"*+ 

and    divisor    as    once    before,    and 

therefore  must  have  the  same  quotient  figure  and  remainder;  and 
this  remainder,  with  a  cipher  annexed,  will  yield  another  quotient 
figure  and  remainder,  like  a  previous  one,  and  so  on  continually. 

ART.  162.  These  interminate  decimals  on  this  account 
have  received  the  name  ^f  circulating  or  recurring"  deci- 
mals; also,  repeating  or  periodical  decima.lt, 

The  figure,  or  s£t  of  figures,  which  is  constantly  re- 
peated, is  called  the  repetend,  which  signifies  to  be  repeated. 

Such  decimals  are  expressed  by  placing  a  dot  over  the 
first  and  last  figures  of  the  first  repetend,  and  omitting 
those  which  follow;  thus,  !  =  .6666+  =  .6  and  \  = 
.l4285t  =  .142857142857+ 

ART.  163.  A  circulate  or  circulating  decimal  has  one  or 
more  figures  constantly  repeated  in  the  same  order. 

A  repetend  is  the  figure  or  set  of  figures  repeated. 


RBVIEW. — 153.  If  tho  marked  figure  of  the  dividend  is  reached  after 
I'he  first  multiplication,  what  should  bo  done?  If  the  dividend  has  not 
tho  proper  plsioo  to  bo  marked,  whnt  should  bo  done?  What  directions 
are  to  bo  observed  in  carrying  tons  from  the  rejected  figures  ?  Solv« 
the  example.  Prove  the  rale.  160.  What  is  Case  37  The  rule? 
irate  and  prove  it. 
11 


122  HAY'S   HIGHER   ARITHMETIC. 

A  pure  circulate  has   no  figures  but  the  rcpctcnd;   as, 
.5  and  .124 

A  mfced  circulate  has  other  figures  before  the  rcpctcnd 
ts,  .208^  and  .31247 

A  simple  rcpetend  has  one  figure;  as,  .4 

A  compc'.nd  rcpctcnd  has  two  or  more  figures;  as,   .51) 

Similar  rcpetends  begin  at  the  same  place;  as,  . 35023 J 
and  .0178,  which  both  begin  at  thousandths. 

Dissimilar  rrpctends  begin  at  different  places;  as,  .205 
and  .312408 

Similar  and  conterminous  repetcnds  begin  and  end  al 
the  same  places;  as,  .5039t  and  .42018 

ART.  164.  Any  terminate  decimal  may  be  considered  a 
circulate,  its  rcpetend  being  ciphers;  as,  .35  =  .350  = 

.350000.  .Any  simple  repctcnd  may  be  made  compound, 
and  any  compound  repctcnd  still  more  compound,  by 
taking  in  one  or  more  of  the  succeeding  rcpctcnds;  as,  .fy 

=  .33333,  and  .0502  =  .0502G2,  and  .257=. 257257257 
When  a  rcpctcnd  is  thus  enlarged,  be  careful  to  take  in 
no  part  of  a  rcpetend  without  taking  the  whole  of  it; 
thus,  if  we  take  in  2  figures  in  the  last  example,  the 
result,  .25725,  would  be  incorrect,  for  the  next  figure 
understood  being  7,  shows  that  25/25  is  not  repeated. 
A  rcpetend  may  be  made  to  begin  at  any  lower  place  by 
carrying  its  dots  forward,  each  the  same  distance;  thus, 

.5  =  . 555,  and  .2941  =  . 29414,  and  5.1830=5.183083 
Dissimilar  repctends  can   bo  made   similar,  by  carrying 
the  dots  forward  till  they  all  begin  at  the  same  place,  as 
the  one  furthest  from  the  decimal  point. 

Similar  repctends  may  be  made  conterminous  by  enlarg- 
ing the  rcpcicnds  until  they  all  contain  the  same  number 
of  figures.  This  number  will  be  the  least  common  mul- 
tiple of  tbs  numbers  of  figures  in  the  givn  ropofcnds. 
For,  suppose  one  of  the  rcpctcnda  to  have  2,  smother.  3, 
another,  4,  and  the  last,  0  figures;  in  enlarging  the  first, 
figures  must  be  taken  in,  2  at  a  time;  ami  in  tho  others. 
3,  4,  and  0  at  a  time.  The  number  of  figures  which  wny 
be  taken  in  2,  3,  4,  and  6  at  a  time,  is  a  common  inul- 


REDUCTION   OF   CIRCULATES.  123 

tiplc  of  these  numbers,  and  the  least  common  multiple  in 
preferred  for  convenience. 


REDUCTION  OF  CIRCULATES. 

CASE  I. — TO  REDUCE  A  PURE  CIRCULATE  TO  A  COMMOH 
FRACTION. 

ART.  165.  RULE. —  Write  the  repelend  for  the  numerator 
and  for  the  denominator  take  as  many  Q's  as  there  are  figure* 
in  the  repetend. 

DEMONSTRATION. — Take  the  pure  circulate  .450  =  .456456456, 
&c.,  and  remove  the  decimal  point  to  the  right  over  one  repetend ; 
the  result  456.456456,  &c.  =  456 . 4*56  is  1000  times  .456,  (Art.  144); 
hence,  the  part  456  =  999  times  .456 ;  and  .4*56  =  li  §  =  i  3  *,  which 
agrees  with  the  rule. 

NOTE. — If  the  repetend  begins  before  the  decimal  point  at  some 
place  of  whole  numbers,  carry  the  dots  forward  until  the  repetend 
begins  at  the  tenths'  place,  and  then  apply  the  rule;  thus,  25.6  = 
25.025  =  25§  §f. 

CASE  II. — TO  REDUCE  A  MIXED  CIRCULATE  TO  A  COMMON 
FRACTION. 

ART.  166.  RULE. — Subtract  the  figures  which  precede  the 
repetend  from  the  whole  circulate  for  tlie  numerator;  for  tfte  de- 
nominator take  as  many  (J's  as  there  are  figures  in  the  repelend, 
with  as  many  ciphers  annexed  as  there  are  decimal  figures  before 
the  repetend. 

Change  .821437  to  its  equivalent  common  fraction. 

OPERATION. 

DEM.— The  work,  QOIAQ'T         eoi<37 

by  the  rule  in  Case  '8'  L4d/  =  'O*1**! 

1,  resnlts  in  _  821^11  _  821  X  999  +  437 

821437-821  ~  IQQQ  999000 

821(1000— 1)  + 437 

for  the  answer,  and  — i — ' = 

this  is  what  would  999000 

be  got,  at  once,  if  the   821000  —  821  +  437      821437—821 

SiltatdTo^Uh  "          999000  999000 

all  other  mixed  cir-  820616     102577 


culate9' 


999000~124875 


RAY'S  HIGHER  ARITHMETIC. 


REDUCE  TO  COMMON  FRACTIONS, 


1.  .3     ....  =i 

2.  .05  ....  =TV 

3.  .123     .    .    .  =3Ys 

4.  2.6S     .    .    .  =2T7T 
6.  .31  ....  =if 

6.  .0216    .       .       .  =181, 

7.  48.1 


8.  1.001  .    .  •- 

9.  .138    .    .  .    =36« 

10.  2083    .    .  .    =& 

11.  85.7142  .      =85? 

12.  .063492  .  .    =  g4,- 

13.  .4476190  .  =  T4o75 

14.  N. 09027     .  •  =TW 

ART.  167.    Circulates  may  bo  added,  subtracted,  multi 
plied,  or  divided,  by  this 

GENERAL  RULE  FOR   CIRCULATES. 

Reduce  the  circulates  to  common  fractions,  and  perform  on 
them  the  operation  required. 

REMARK. — Circulates  may  be  carried  forward  far  enough  to 
avoid  any  sensible  error  in  the  result,  and  then  treated  as  other 
decimals.  They  can  be  added,  subtracted,  multiplied,  and  divided, 
without  this  preparation;  as  will  now  be  explained. 


ADDITION    OF    CIRCULATES. 

ART.  168.  RULE. — Malta  the  repetends  similar  and  conter- 
minous, if  they  be  not  so;  add,  and  point  off  as  in  ordinary 
decimals,  increasing  the  right-hand  column  by  the  amount  if 
any,  which  would  be  carried  to  it  if  the  circulates  were  continued ; 
then  make  a  repetend  in  the  sum,  similar  and  conterminous  with 
those  above. 

Add  .256,  5.3472,  24.815,  and  .9098 

DEMONSTRATION. — Make  the  cir-  .2566666666 

culates  similar  and  conterminous,  as  ^Q4.>79'79I7O7O 

directed  in  Art.  164.     The  first  column 

of  figures  which  would  appear,  if  the       24.8158158158 
uirculates  were  continued,  are  the  same  .9098000000 

as  the  first  figures  of  the  repetends, 
6,  7,  1,  0,  whose  sum,  14,  gives  1  to  be  S 
carried  to  the  right  hand  column.  Since  the  last  six  figures  in 
each  number  is  a  repetend,  the  last  six  figures  of  the  sum  is  also  • 
repetend 


SUBTRACTION  OP  CIRCULATES.  135 

REMARK. — In  finding  the  amount  to  be  carried  to  the  right 
hand  column,  it  may  be  necessary,  sometimes,  to  use  the  two  suc- 
ceeding figures  in  each  repetend. 

1.  Add  .455,  .06$,  .32t,  .945  Ant.  1.796 

2.  Add  3.04,  6.456,  23.3$,  .248       AM.  33.1334 

3.  Add  .£5,  .104,  .61,  and  .5635  AM.  1.536 

4.  Add  1.03,  .25?,  5.64,  28.0445'245    AM.  34.3? 

5.  Add  .6,  .13§,  .05,  .0972,  .0416  Ans.  1. 

6.  Add  9.2liOt,  .65,  5.004,  3.5622     AM.  18.43 

7.  Add  .2045*,  .69,  and  .25  Ans.  .54 

8.  Add  5.0*776,  .24,  and  7.124943  AM.  12.4 

9.  Add  3.4884,  1.63?,  130.81,  .066  AM.  136. 06 


SUBTRACTION  OF  CIRCULATES. 

ART.  169.  RULE.  —  Make  the  repetends  similar  and  conter- 
minous, if  they  be  not  so  ;  subtract  and  point  off  as  in  ordinary 
decimals,  carrying  one,  however,  to  the  right-hand  figure  of  the 
subtrahend,  if  on  continuing  the  circulates  it  be  found  necessary  ; 
then  make  a  repetend  in  the  remainder,  similar  and  conterminous 
with  those  above. 

Subtract  9.3l56  from  12.9021 

DEMONSTRATION.  —  Prepare  the  num-  12.90212121 
bers  for  subtraction.  If  the  circulates  were  9.  31561561 
continued,  the  next  figure  in  the  subtrahend 


, 

(5)  would  be  larger  than  the  one  above  it        3. 
(2);  therefore,  carry  1  to  tho  right-hand  figure  of  the  subtrahend. 

REMARK.  —  It  may  be  necessary  to  observe  more  than  one  of  th« 
ruccecding  figures  in  tho  circulates,  to  ascertain  whether  1  i?  to  b« 
carried  to  the  right-hcnd  figure  of  the  subtrahend  or  not 

1.  Subtract  .0674  from  .26      .....     =  .25$ 

2.  Subtract  9.  Ofr  from  15.  35465"       .'     .     .     =  6.2o 

3.  Subtract  4.51  from  18.23673      .     .     .=13.72 

4.  Subtract  37.0l2«  from  100.73    ,  .  =  63.  tl 


126  RAY'S   HIGHER  ARITHMETIC. 

5.  Subtract  8. 2f  from  10.0563    .    .=1.7836290 

6.  Subtract.  190.476  from  199.6428571       =  9.l6 


MULTIPLICATION  OF  CIRCULATES. 

ART.  170.  RULE. — If  only  one  of  tJie  numbers  be  a  circulate, 
make  it  the  multiplicand,  and  perform  the  work  as  in  ordinary 
decimals,  carrying  to  the  right  hand  figure  of  each  product,  the 
amount  that  would  be  necessary  if  the  multiplicand  were  continued 
further;  make  the  repetends  in  the  partial  products  similar  and 
•ionlerininous,  and  add  according  to  the  rule  already  given. 

If  the  multiplier  have  a  repetend,  reduce  it  to  a  common  frac- 
tion and  add  in  the  result  obtained  by  using  this  fraction. 

Multiply  .3754  by  17. 43 

SOLUTION. — In  forming  the  pa r-  .3754 

tial  products,   carry   to   the    right-     -JK  4  A  _  -j  w  ±\ 
band   figures  of  each   respectively, 


the  numbers  1,  3,  0,  arising  from  the  .1501777 

multiplication  of  the  figures  that  do  2.6281111 

not  appear.     The  rcpetend  of  the  q  *  K.AAA  A\ 

multiplier  being  equal  to  I ,  |  of  the  '  °  *  *  .  * 

multiplicand  is  12oi48,  whose  figures  1  ~  5 1 4  8 

aro  set  down  under  those  of  the  mul-  6.5452481  Am. 

tirlicand  from  which  they  were  ob- 
is'ned.     Point  the  several  products,  carry  them  forward,  until  thi»il 
rr  petends  are  similar  and  conterminous,  and  add  for  answer. 

1.  4.735x7.349 =34.800113 

2.  .Ot06fx.9l32 =.06(5065 

3.  7^4.32X3.456 =2469. 17381 4 

4.  16.204  X  32. *75 =530.810446 

6.  19.0t$X.2083 =3.973l& 

6.  10.0ol2x4.263     ....    =42.85-188033 

7.  3.t543x  4.7157 =  17.7045083 

6.    1.256784x6.42081  .  =8.069o83206 


DIVISION   OP  CIRCULATES.  127 

DIVISION  OF  CIRCULATES. 

ART.  171.  RULE.  —  Make  the  repetends  similar  and  conter- 
minous. Subtract  from  each  circulate  the  figures  preceding  its 
repetend,  and  use  the  remainders  for  the  dividend  and  divittr 
respectively,  omitting  the  dots. 

NOTE.  —  If  the  divisor  is  not  a  circulate,  it  will  be  shorter  to 
divide  as  in  ordinary  decimals,  bringing  from  the  dividend  the 
figures  of  the  rcpetend  instead  of  ciphers,  to  continue  the  division. 

Divide  2.6o3  by  1.8 

FIRST   OPERATION.  SECOND    OPERATION. 

1.800  2.653  1.8)2.65§(1.4t4d 

18         2C5  85  / 


__ 
1620)2388(1.4f4(5 

708 
1200 

GGO 
1200 

DEMONSTRATION.  —  The  remainders  1620  and  2388,  in  the  1st 
operation,  are  the  numerators  of  ihe  common  fractions  to  which  the 
circulates  are  equivalent,  (Art.  1G6);  and  as  they  have  the  same 
denominators  (each  being  as  many  9's  as  there  are  figures  in  the 
repeteml,  with  as  many  ciphers  annexe.l  as  there  are  decimal 
figures  before  the  repetend,  Art.  166),  dividing  these  numerators 
will  give  the  same  quotient  as  if  the  fractions  themselves  were  used, 
and  therefore  the  dots  may  be  omitted. 

EXAMPLES  FOR  PRACTICE. 

1.  .?&•*-.!        ..........     =6.S1 

2.  Sl^Dl-f-lY    ........       =  3.02& 

3.  S81.559887&-5-94  .....  =7.2506371 

4.  90.  5^03749-^6.  7o4    .....    =13.4ol 
6     11.6G8735402-s-.24o  .....      =  45.1$ 

6.  9  5330GG3997-f-6.£l7     .....    =1.53 

7.  3.500«>91358024-f-T.684     .    .         .      =  .4o 
Multiplication    and    division   of  circulates   can   be   fre- 

quently  performed   with   advantage    by   the   general  rula 
for  circulates,  (Art.  167). 


128  RAY'S   HIGHER   ARITHMETIC. 

ART.  172.  There  is  a  short  method  of  converting  a  com- 
mon fraction  into  a  circulating  decimal,  when  the  denomi- 
nator is  a  prime  number,  which  is  worth  considering. 

By  actual  division,  ?  =  .142857,  which  has  this  prop- 
erty, viz : 

1st  Property.  The  number  of  figures  in  the  repctend  ia 
either  one  less  than  the  denominator,  or  a  half,  a  third,  or 

ifjme  other  exact  part  of  this  one  less;  thus,  77T  =  63,  the 
cumber  of  figures  in  the  repetend  (2)  being  £  of  10. 

This  is  true  of  any  fraction,  -whose  denominator  is  a  prime  number 
other  than  5.  If  the  number  of  figures  in  the  repetend  is  one  less  than 
the  denominator,  two  other  properties  are  observed. 

2d  Property.  Each  figure  in  the  first  half  of  the  repe- 
tend added  to  the  corresponding  figure  in  the  last  half, 

makes  9;  thus,  in  .14285t,  1  +  8,  4  +  5,  2  +  7,  each 
equals  9. 

3d  Property.  The  same  repetend  serves  for  all  fractions 
having  the  same  prime  denominator,  whatever  be  their 
numerators,  by  starting  at  different  places ;  thus,  ^  = 

.142857,  f  =  .285714,  4  =  .428571. 

Convert  33  into  a  circulating  decimal. 

SOLUTION. — First,  by  actual  division,  ^  =  .0434782fi5  ;  instead 
of  363,  put  G  times  the  value  of  ^g  just  given,  viz:  .0434782G0869^f ; 
having  12  figures  without  repeating,  the  repetend  must  be  of  22 
figures  by  property  first:  obtain  the  other  figures  according  to  the 
2d  property,  by  subtracting  each  of  the  first  11  from  9;  thus, 
^  =  .6434782008695652173913;  to  get  Af  use  this  same  repetend, 
and  to  ascertain  the  starting  place,  multiply  the  first  3  or  4  figures 
by  10;  thus,  .0434  multiplied  by  1C,  gives  .G944,  showing  that,  w« 
must  commence  at  the  eleventh  figure;  doing  BO,  the  value  of  ^§  = 
6950521739130434782608. 

CONVERT   INTO   CIRCULATING   DECIMALS, 

S  5  11  8  7          14        24         49  3  10        fi?         65 

iV     T7>     T¥J     3U>     3T»     Sl>     "4T>     g?>     7oT>     *S)     71>     89' 


XI.   COMPOUND  NUMBERS. 

ART.  173.    A  simple  number  is  of  one  denominatiop  :  MS 
Ho  dollars  ;  2000  bushels  ;  4  apples. 


COMPOUND  NUMBERS.  129 

A  compound  number  consists  of  several  simple  num- 
bers of  different  denominations;  as,  16  dollars  75  cents ; 
3  yards  2  feet  8  inches. 

Compound  numbers  are  often  called  denominate  num- 
bers ;  they  are  used  for  measures,  weights  and  money. 

LONG  OR  LINEAR  MEASURE 

ART.  174.  Is  used  for  measuring  distance,  also  the 
7ength,  breadth,  and  hight  of  bodies,  called  their  tineat 
dimensions. 

TABLE. 

12  inches,  (in.) make  1  foot,  (ft.) 

3     ft : 1  yard,  (yd.) 

5^  yd.  or  16£  ft. 1  rod,  (rd.) 

40  rd 1  furlong,  (fur.) 

8     fur. 1  mile,  (mi.) 

REMARK. — The  rod  is  sometimes  called  pole  or  perch. 

NOTES  . — 1.  4  in.  =  1  hand,  used  in  measuring  the  hight  of 
horses ;  9  in.  =  1  span ;  3  feet  =  1  pace ;  18  in.  =  1  cubit ;  3  in. 
=  1  palm. 

2.  The  scale  used  by  carpenters  has  the  foot  divided  into  12  in.: 
each  inch  divided  into  12  equal  parts,  called  lines;  each  line  into  12 
equal  parts,  called  seconds;  and  each  second  into  12  equal  parts, 
called  thirds.  The  inch  in  these  scales  is  also  divided  into  eighths 
and  sixteenths,  and  tenths. 

MARINERS'   MEASURE 

ART.  175.  Is  a  kind  of  Long  Measure  used  in  esti- 
mating distances  at  sea. 

0  feet „ make  1  fathom. 

120  fathoms 1  cable-length. 

880  fathoms,  or  7  3  cable-lengths 1  mile. 

NOTE. — 1  nautical  league  =  3  equatorial  miles  =  3.45771  statute 
miles.  60  equatorial  miles  =  69.1542  statute  miles  =  1  equatorial 
degree;  360  equatorial  degrees  =  1  great  circle,  or  circumference 
of  the  earth. 

REVIEW. — 173.  What  is  a  ftimplo  number?  a  compound  number? 
Giro  examples.  What  are  compound  numbers  often  called  ?  What  ar» 
they  used  for?  174.  What  is  long  measure  used  forT  Repeat  tho  table. 
175.  What  is  mariners'  measure?  Repeat  tho  table. 


130  RAY'S   HIGHER  ARITHMETIC. 


SURVEYORS'  AND  ENGINEERS'  MEASURE 

ART.  176.  Is  a  kind  of  long  measure,  used  in  laying 
out  roads,  and  running  the  boundaries  of  land. 

100  links  (Ik.) make  1  chain,  (ch.) 

80  chains  (ch.) 1   mile,  (mi.) 

NOTES. — 1.  The  chain    is   called  surveyors'   or  Gunter's   chain, 
from  its  inventor,  and  is  4  rods,  or  CO  feet  in  length.     As  it  consists 
of  100  links,  each  link  must  be  7.92  in.  long;  hence,  to  change  these 
denominations  to  the  ordinary  linear  measure,  recollect,  that 
1  chain  =  4  rd.  or  GO  feet. 
1   link  =  7.9'2  in. 

2.  Since  each  link  is  j^  of  a  chain,  the  number  of  links  can  be 
written  as  decimal  hundredths  with  the  whole  chains,  as  2.56  chains 
=  2  chains  56  links. 

REMARK. — Inches  and  yards  are  not  used  in  the  last  two  kinds 
of  measurement. 

CLOTH   MEASURE 
ART.  177.    Is  a  kind  of  long  measure  used  for  dry  goods. 

2  J   in make  1  nail  (na.), 

4     na.  or  9  in 1  quarter,  (qr.) 

4     qr 1  yd. 

NOTES. — 1.  1  ell  Flemish  =  3  qr.  or  |  yd.;  1  ell  English  = 
6  qr.  or  ij  yd.;  1  ell  French  =  6  qr.  or  l|  yd. 

2.  At  the  custom-houses,  the  yard  only  is  used,  being  divided 
into  tenths  and  hundredths;  in  mercantile  transactions,  the  yard 
is  the  unit,  and  the  fractional  parts  employed  are  quarters,  eighths, 
sixteenths,  and  half-sixteenths. 

ART.  178.  The  standard  of  all  our  linear  measure  is 
the  yard,  being  identical  with  the  imperial  yard  of  Great 
Britain,  which  was  determined  as  follows : 

By  accurate  experiment  at  London,  the  length  of  a 
pendulum  was  ascertained,  which,  in  a  vacuum,  at  the 
level  of  the  sea,  vibrated  86400  times  in  a  mean  solar 

R  K  v  i  E  w  . — 17i>.  Whiit  is  surveyors'  men.-ure?  Repent  the  Inlilo. 
What  is  tho  chain  called?  Why?  How  long  is  it?  lluw  long  is  H  link? 
Why?  How  nro  chains  and  links  written  together?  17T.  What  is  cloth 
measure?  Repeat  tho  table-.  What  is  an  ell  Flemish  ?  An  oil  English? 
An  oil  French  ? 


COMPOUND  NUMBERS.  13] 

day,  or  once  every  second.  This  pendulum  was  divided 
into  391393  equal  parts,  and  3COOOO  of  these  were 
taken  to  be  a  yard,  the  pendulum  itself  then  being 
39.1393  inches  long. 

REMARK. — Linear  measurement,  as  being  especially  necessary, 
was  used,  and  to  a  certain  degree  fixed,  at  an  earlier  period  than  the 
measures  of  volume  and  weight,  which  have  therefore  been  made  to 
depend  upon  and  be  verified  by  the  former. 

The  standards  of  long  measure  were  at  first  very  imperfect,  being 
derived  from  different  parts  of  the  human  body,  such  as  a  finger- 
joint,  finger,  hand,  span,  cubit  or  fore-arm,  and  yard  or  whole  arm; 
but  as  commerce  increased,  and  convenience  demanded  a  change, 
they  were  rendered  precise  and  uniform. 

The  ancient  yard  of  Great  Britain  is  said  to  have  been  determined 
by  the  length  of  the  arm  of  King  Henry  I. 

SQUARE  OR  SURFACE  MEASURE 

AHT.  179.  Is  used  in  estimating  the  contents  of  land, 
painters'  and  plasterers'  work,  and  other  surfaces. 

A  square  is  an  even  surface,  bounded  by  four  straight 
lines  or  sides.  Each  side  is  perpendicular  to  two  others. 

The  size  or  name  of  any  square  depends  upon  that  of  its  side; 
a  square  inch  is  a  square,  whose  side  is  an  inch  long;  a  square 
foot,  one  whose  side  is  a  foot  long,  and  so  on. 

AHT.  180.  The  unit  by  which  all  surfaces  are  measured 
is  a  square,  whose  side  is  a  linear  inch,  foot,  yard,  rod  or 
mile;  and  the  size  of  any  surface  will  be  the  number  of 
times  it  contains  this  unit. 

The  simplest  surface  is  a  rectangle,  which  is  an  even  surface, 
having  four  straight  lines  for  sides,  each  opposite  pair  being  equal, 
and  perpendicular  to  the  other  pair.  The  ceiling  and  sides  of  a 
room,  and  sheets  of  paper,  arc  examples  of  rectangles. 

If  the  length  and  breadth  of  a  rectangle  are  the  same,  the  sides 
are  all  equal,  and  it  is  a  square. 

The  size,  or  area  of  a  rectangle,  being  the  number  of 
square  measuring  units  it  contains,  can  be  ascertained  aa 
follows: 

REVIEW.— 178.  What  is  tho  standard  unit  of  length  in  the  United 
States?  How  is  it  determined?  What  measures  of  length  were  used 
at  first?  179.  What  is  square  measure?  What  is  a  square?  A  «q, 
inch?  A  B(\.  ft.?  A  sq.  yd.?  ISO.  What  is  tho  unit  of  measure  for  till 
surfaces  ? 


132  RAY'S   HIGHER   ARITHMETIC. 


Take  a  rectangle  4  inches  long  by  3  inches 
wide.  If  upon  each  of  the  inches  in  the  length, 
a  square  inch  be  conceived  to  stand,  there  will 
be  a  row  of  4  square  inches,  extending  the  whole 
length  of  the  rectangle,  and  reaching  1  inch  of 
its  width.  As  the  rectangle  contains  as  many  such  rows  as  there 
are  inches  in  its  width,  its  area  must  be  equal  to  the  number  of 
square  inches  in  a  row  (4)  multiplied  by  the  number  of  rows  (3),  = 
12  square  inches ;  hence,  to  find  the  area  of  a  rectangle, 

RULE. — Multiply  the  number  of  linear  units  in  the  length  by 
the  number  of  linear  units  in  the  breadth,  after  expressing  them 
in  the  same  denomination.  The  product  will  be  the  area  in 
square  units  of  tJie  same  denomination. 

It  is  easy  to  determine  the  number  of  sq.  inches  in  a 
Bq.  foot,  of  sq.  feet  in  a  sq.  yd.,  and  so  on;  for,  since  1 
sq.  foot  is  12  inches  long  by  12  inches  wide,  it  must 
contain  12  X  12  =  144  sq.  in.;  and  since  1  sq.  yd.  is  3 
feet  long  by  3  feet  wide,  it  must  contain  3  X  3  =  9  sq. 
feet;  and  since  1  sq.  rod  is  5Ayd.  long  by  5Ayd.  wide, 
it  must  contain  63  X  5A  =  30j  sq.  yd. 

144  square  inches  (sq.  in.)  make  1  square  foot,  (sq.  ft.) 

9  sq.  ft 1  square  yard,  (sq.  yd.) 

30]  sq.  yd 1  square  rod,   (sq.  rd.) 

LAND  MEASURE 

ART.  181.  Is  a  kind  of  surface  measure,  used  to  ox- 
press  the  contents  of  land. 

40  perches  (P.)  make  1  rood,  (R.) 

10  sq.  chains,  or  4  11 1  acre,  (A.) 

G40  A, 1  eq.  mile,  (sq.  mi.) 

NOTE.  —  Since  1  chain  =  4  rods,  1  sq.  chain  =  4X4  =  lG  sq 
rods.  Since  links  are  written  as  decimal  lumdreths  of  a  chain, 
chains  and  links  can  be  considered  as  chains  only;  thus,  7  chains 
and  9  links  =  7.09  chains,  and  the  area  of  a  square  whose  side  is 
7.09  chains,  will  be  expressed  in  square  chains  and  a  decimal,  ai 
follows:  7.09  X  7.09  =60.2681  square  chains;  there  is  no  necessity 
then,  of  using  the  denominations  link  or  square  link  in  practice. 

REVIEW. — 180.  What  is  a  rectangle?  What  is  the  rule  for  the  area 
of  a  rectangle?  Prove  it.  How  many  square  inches  in  a  square  foot? 
Why  T  How  many  square  feet  in  a  square  yard  ?  Why  ?  How  many 
iquare  yards  in  a  square  rod?  Why?  Repeat  the  table. 


COMPOUND  NUMBERS.  133 


CUBIC  OE  SOLID   MEASURE 

ART.  182.  Is  used  to  measure  the-  bulk  of  stone,  tim- 
ber, masonry,  and  other  solid  work;  to  find  tlic  contents 
of  cellars,  and  to  verify  measures  of  capacity. 

A  cube  is  a  solid,  bounded  by  six  equal  squares  or 
faces,  each  opposite  pair  of  which  is  perpendicular  to  tho 
other  four.  Its  length,  breath  and  hight,  then,  are  all 
equal,  and  each  is  called  the  side  of  the  cube. 

The  size  or  name  of  any  cube,  like  that  of  a  square,  depends 
upon  its  side,  as  cubic  inch,  cubic  foot,  cubic  yard. 

ART.  183.  The  unit  by  which  all  solids  are  measured 
is  a  cube,  whose  side  is  a  linear  inch,  foot,  &c.,  and  their 
size  or  solidity  will  be  the  number  of  times  they  contain 
this  unit. 

The  simplest  solid  is  the  rectangular  solid,  which  is  bounded  by 
six  rectangles,  called  its  faces,  each  opposite  pair  being  equal, 
and  perpendicular  to  the  other  four;  a  bar  of  soap,  a  candle-box, 
are  rectangular  solids.  If  the  length,  breadth,  and  hight  are  the 
same,  the  faces  are  squares,  and  the  solid  is  a  cube. 

The  size,  or  solidify  of  any  rectangular  solid  is  found 
as  we  obtain  the  area  of  a  square  (Art.  180.) 

Suppose  the  rectangle  4  inches  long  by  3  inches  \vide,  in  Art.  180, 
to  be  tno  bottom  or  lower  base  of  a  rectangular  solid,  its  upper  face 
being  01  the  same  dimensions,  and  its  hight  5  inches. 

If  upon  each  of  the  12  square  inches  in 
the  IOWCA*  base,  a  cubic  inch  be  conceived 
to  stand,  there  will  be  a  section  of  12  cubic 
inches  covering  the  whole  bottom  of  the 
solid,  and  reaching  1  inch  of  its  hight;  as 
the  solid  contains  as  many  such  sections 
as  there  are  inches  in  hight,  its  solidity 
must  be  equal  to  the  number  of  cubic 
inches  in  a  section  (12)  multiplied  by  the 

number  of   sections   (5)  or  60  cubic  inches.     But  the  number  cf 
cubic   inches   in  a  section  is  the  same  as  the  number  of   square 

REVIEW. —  1 81.  Whnt  i?  land  measure?  Repeat  the  table.  How  many 
square  chains  in  nn  arro?  Why?  Wliy  do  we  not  u?e  square  linkf? 
182.  What  is  cubic  measure?  Whut  is  a  cube?  A  cubic  inch?  A  cubic 
foot?  A  cubic  yard?  183.  What  is  the  unit  fur  all  solids?  What  is  a 
rectangular  solid  ? 


134  RAT'S  HIGHER   ARITHMETIC. 

inches  in  (he  b.ase,  (12,)  and  this  again  is  equal  to  the  number  of 
linear  inches  in  the  length  (4),  multiplied  by  the  number  in  tha 
width,  (3);  hence, 

TO  FIND   TUB   SOLIDITY   OF   A   RECTANGULAR    SOLID, 

RULE. — Multiply  the  length,  breadth,  and  hiyht  together,  aftei 
expressing  them  in  the  same  denomination;  the  product  will  be 
ike  solidity  in  cubic  units  of  the  same  denomination. 

It  is  easy  to  determine  the  number  of  cubic  inches  in 
a  cubic  foot,  and  of  cubic  feet  in  a  cubic  yard. 

For,  since  1  cubic  foot  is  12  inches  long,  12  inches 
wide,  and  12  inches  high,  its  solidity  \vill  be  12  X  12  X 
12  =  1728  cubic  inches;  and,  since  1  cubic  yard  is  3 
feet  long,  3  feet  wide,  and  3  feet  high,  its  solidity  will  be 
3  X  3  X  3  =  27  cubic  feet;  hence, 

1728  cubic  inches  (cu.  in.)  make  1  cubic  foot    (cu.  ft.) 
27  CH.  ft 1  cubic  yard  (cu.  yd.) 

NOTES. — 1.  1  tun  of  round  timber  =  40  cu.  ft.;  1  tun  of  hewn 
timber  =  50  cu.  ft.;  1  tun  of  shipping  =  42  cu.  ft. 

2.  1  cord  of  wood,  8  ft.  long,   4  ft.  wid3,  and  4  ft.  high,    contains 
8  X  4  X  4  =  128  cu.  ft. ;  1  cord  foot  or  foot  of  -wood,  1  ft.  long,  4  ft. 
wide,  and  4  ft.  high,  contains  1  X  4  X  4  =  1G  cu.  ft. 

3.  1  reduced  foot,  plank  measure,  1  ft.  long,    1  ft.  wide,  and  1  in. 
thick,  contains  12  X  12  X  1  =  144  cu.  in.;  all  planks  nnd  scantling 
less   than   an   inch   are  reckoned  1  inch  thick;  but,  if  more  than 
1    inch    thick,  allowance   must  be  made  bj   multiplying   by   that 
dimension. 

4.  1  perch  of  masonry,  1  rod  long,  1  ft.  high  and  li  ft.  thick, 
contains  IGi   X  1  X  U  =  V  X  2  =  V  =  24f  cu.  ft*  which  is 
usually  taken  25  cubic  feet  in  practice. 

TROY  OR  MINT  WEIGHT 

ART.  184.  Is  used  for  weighing  gold,  silver,  jewels,  in 
testing  the  strength  of  spirituous  liquors,  in  philosophical 
•xpcriments,  and  in  comparing  different  weights. 

24     grains  (  gr.) make  1  pennyweight,  ( pwt.) 

20    p\vt 1  ounce,  (  oz.) 

1*2     oz 1   pound,   (lb.) 

R  •  v  i  K  w.—  133.  What  i;<  the  rule  for  its  solidify  ?  Prove  it  How 
many  cubic  inches  in  a  cubic  foot?  Why?  How  many  cubic  feet  in  a 
oubic  yard  ?  Why?  Repeat  the  table.  What  is  a  cord  of  wood?  A  perch 
of  masonry  ?  Bow  many  cubic  feot  in  each  T 


COMPOUND  NUMBERS.  133 

REMARKS. —  1.  The  Troy  pound  is  the  standard  of  weight  in  the 
D.  S.  Mints;  it  is  identical  with  the  imperial  Troy  pound  of  Great 
Britain,  which  contains  6760  grains,  252.458  of  which  are  equal 
in  weight  to  a  cubic  inch  of  distilled  water  when  the  barometer  is 
30  inches,  and  the  thermometer  (  Fahrenheit's)  02°. 

2.  The  name  Troy  is  supposed  by  some  to  be  derived  from  Troyes, 
a  city  of  France,  where  this  weight  was  first  introduced  from  the 
East,  during  the  Crusades;  and  by  others  from   Troy  Novant,  the 
ancient  name  of  London.     The  ounce  Troy  is  the  only  denomination 
used  ut  the  mint,  all  amounts  of  gold  and  silver  being  expressed  in 
it  and  its  decimal  divisions. 

3.  The   Troy   pound  is   equal   to  the  weight  of  22.794422  cubic 
inches  of  distilled  water,  at  the  temperature  of  39°.83  Fahrenheit, 
the  barometer  being  30  inches. 

4.  The  grain  was   originally  fixed  by  taking   a  grain  of  wheat 
from  the  middle  of  the  ear,  and  thoroughly  drying  it;  at  first,  32  of 
these  grains  made  a  pennyweight,  but  afterward  24. 

6.  The  pennyweight  was  the  weight  of  the  silver  penny  in  use  at 
that  time,  and  is  marked  (pwt.)  from  (p.)  for  penny,  and  (  wt.)  for 
weight. 

DIAMOND  WEIGHT 

ART.  185.  Is  used  for  weighing  diamonds  and  other 
precious  stones. 

16    parts make  1  carat  gniin  =    .8  Troy  grain. 

4    carat  grains 1  carat.  =  3.2    do.      do. 

REMARK. — The  carat  in  this  table  is  an  absolute  weight,  and  must 
be  carefully  distinguished  from  the  same  word  used  in  speaking  ot 
the  fineness  of  gold,  for  then  it  indicates  the  proportion  of  pure  gold 
in  a  mass. 

APOTHECARIES'  WEIGHT 

ART.  186.  Is  chiefly  used  in  mixing  medical  prescrip- 
tions. 

20  grains  (gr.) make  )   scruple,  O.) 

3  9 1  dram,     (3.) 

8  3 1  ounce,    (3.) 

12  £ 1  pound,  (lb.) 

The  pound,  ounce,  and  grain,  of  this  weight,  are  the  same  as  those 
of  Troy  weight;  the  pound  in  each  contains  12  o»..  =  5700  gr. 

RKVIEW. — 184.  What  is  Troy  Weight?  Repeat  the  tnMe.  What  ii 
said  of  the  Troy  pound  ?  The  name  Tn.y  ?  What  denoiuinuli*>n  onljr  it 
Mod  at  the  mint?  What  ia  said  of  the  grain  ? 


136  RAY'S   HIGHER   ARITHMETIC. 


AVOIRDUPOIS  OR  COMMERCIAL  WEIGHT 

ART.  187.  Is  used  in  commercial  transactions  when 
goods  arc  bought  or  sold  by  the  quantity.  Heavy  and 
bulky  articles,  as  groceries,  the  coarser  metals,  drugs,  to,, 
are  weighed  by  it. 

16  drams  (dr.) make  1   ounce,  (oz.) 

16  oz 1  pound,  (Ib.) 

.25  Ib 1  quarter,  (qr.) 

4  qr 1  hundred-weight,  (cwt.) 

20  cwt 1  tun,  (T.) 

NOTES. — 1.  A  stone  =  14  Ib.;  but  a  stone  of  fish,  or  butcher's 
meat  =  8  Ib.,  and  a  stone  of  glass  =  5  Ib.;  a  seam  of  glass  = 

24  stone  =  120  Ib.;  1  pack  of  wool  =240  Ib. 

2.  In  Great  Britain,  the  qr.  =  28  Ib.,  the  cwt.  ==112  Ib.,  the  tun 
=  2240  Ib.     These  values  are  used  at  the  U.  S.  custom-houses  in 
invoices  of  English  goods ;  but  generally  in  this  country  the  qr.  = 

25  Ib.,  the  cwt,  =  100  Ib.,  and  the  tun  =  2000  Ib. 

3.  The  Ib.  avoirdupois  is  equal  to  the  weight  of  27.7274  cu.  in.  of 
distilled  water  at  62°  (Fall.) ;  or  27.7015  cu.  in.  at  39°.83  (Fah.),  the 
barometer  at  30  in.     For  ordinary  purposes,  1  cubic  foot  of  water 
can  be  taken  62  i  Ib.,  or  1000  oz.  avoirdupois. 

4.  The  terms  gross  and  net  are  used  in  this  weight.     Gross  weight 
is  the  weight  of  the  poods,  together  with  the  box,  cask,  or  whatever 
contains  them.     Net  weight  is  the  weight  of  the  goods  alone. 

REMARK. — The  word  avoirdupois  is  from  the  French  avoirs,  rf«, 
pois,  signifying  goods  of  weight.  The  Ib.  avoirdupois  differs  from  the 
Ib.  Troy  or  apothecaries,'  the  former  being  7000  gr.,  the  latter,  each 
57GO  gr.  The  oz.  avoirdupois  also  differs  from  the  oz.  Troy  or 
apothecaries.' 

COMPARISON    OF   WEIGHTS. 

ART.  188.  Since  1  Ib.  av.  =  7000  gr.  Troy,  1  oz.  av. 
=  7»g  of  a  Ib.  av.,  =  T'ff  of  7000  gr.  Troy  =  437A  gr. 
Troy;  and  1  dr.  av.  =  7's  of  an  oz.  av.,  =  T'ff  of  437 A  gr. 
Troy  =  27-£Agr.  Troy;  and  in  a  similar  way  1  oz.  Troy 

HKVIEW. — 185.  llepc.it  the  table  of  diamond  weight.  What  is  said  of 
aoarat?  ISIi.  What  is  apothecaries'  woiirht?  Rjjn'Mt  t!:^  ?:il.!<\  U'ii.it 
denominations  ;in-  iiloiilicul  in  Truy  ami  iipulht'iMrii.'.-'  \vc.'i,ht?  1^7  \Vlia! 
ifl  avoirdupois  weight?  Repeat  the  table.  What  is  s;iiil  of  the  qr.,  o\vt., 
and  tun,  in  Great  Britain  and  the  U.  S.  custom-houses?  What  clseweref 
What  is  said  of  the  Ib.  avoirdupois?  What  is  gross  weight?  Net  weight? 


COMPOUND  NUMBERS.  137 

and  apothecaries'  =  480  gr.  Troy;  and  1  dr.  apothecaries' 
=  60  gr.  Troy.     Hcn^c, 

1  Ib.  avoirdupois =  7000  gr.  Troy  or  apoth. 

1  oz do =  437^  do. ..do do. 

1  dr do =27H  do.. .do do. 

1  Ib.  Troy  or  apoth =  5760  do.. .do.... ....do. 

1  oz.   do do =  480    do.. .do do. 

1  dr.  apothecaries' =  60       do. ..do do. 

This  table  serves  to  convert  denominations  of  one  kind  of  weight 
inlo  those  of  another. 

WINE   Oil   LIQUID   MEASURE 

ART.  189.  Is  used  for  measuring  all  liquids,  except 
ale,  beer,  and  milk. 

4  gills  (gi.) make  1  pint,  (pt.) 

2  pt 1  quart,  (qt  ) 

4  qt 1  gallon,  (gal.)  =  231  cu.  in. 

NOTES.— 1.  3U  gnl.  —  1  barrel  (bbl.);  63  gal.  =  1  hogshead 
(hhd.);  42  gal.  =  1  tierce ;  84  gal.  =  1  puncheon;  126  gal.  =  1  pipe 
or  butt ;  2  pipes  =  1  tun.  These  are  generally  mentioned  as  mea- 
sures, but  are  only  vessels,  and  are  gauged  and  sold  by  the  gallon, 
not  being  of  uniform  capacity.  When  the  contents  of  cisterns, 
wells,  &c.,  are  expressed  in  hhd.  or  bbl.,  they  have  the  values  given 
a"bove. 

2.  In  England,  1  anko,  =  10  gal.;  1  runlet  =  18  gal. 

ART.  190.  The  Mandard  unit  of  liquid  measure  in  the 
U.  S.  is  the  wine  <i<. .ion,  formerly  used  in  Great  Britain, 
containing  231  cu.  in.,  and  which  contains  a  weight  of 
58372.1754  gr.,  =  nearly  8i  Ib.  av.,  of  distilled  water  at 
39°. 83  Fahrenheit  tLo  barometer  at  30  inches. 

A  pint  of  water  is  generally  considered  1  pound. 

ALE    AND   BEER  MEASURE 

ART.  191.  Is  usea  in  measuring  ale,  beer,  and  milk, 
though  milk  is  frequently  sold  by  wine  measure. 

2  pints  (pO  •»•  ;ke  1  quart,  (qt.) 

4  qt 1  gallon,  (gal.)  =  282  cu.  in. 

REVIEW. — 187.  Wba  .es  avoirdupois  mean?  How  docs  the  Ib.  avoir- 
dupois  differ  from  the  Ib  Troy  and  apothecaries  ?  183.  Repeat  the  table 
for  comparison  of  weigh  ISO.  What  is  wine  measure? 

12 


135  RAY'S   HIGHER   ARITHMETIC. 

NOTES. — 1.  1  bbl.  =  3G  gal.;  1  hhd.  =  6-1  gal.  Tbcso  arc  men- 
tioned as  measures,  but  are  only  vessels,  and  are  gauged  and  sold 
by  (he  gallon,  not  being  of  uniform  capacity. 

2.  In  England,  1  firkin  =  9  gal.;  1  kilderkin  =  2  firkins;  1 
bbl.  =  2  kilderkins. 

REMARK.— The  beer  gallon  contains  282 cu. in.,  or  10.179933  Ib.  ay 
f  distilled  water  at  39°.83  Fahrenheit,  the  barometer  at  30  icoues. 

DRY   MEASURE 

ART.  192.  Is  used  for  measuring  grain,  fruit,  vegcta- 
Dleo,  coal,  salt,  &c. 

2  pints  (pt)  make  1  quart,  (qt.) 

8  qt 1  peck,  (pk.) 

4  pk 1  bushel,  (bu.)  =  2150.42  cu.  in. 

NOTES. — 1.  4  qt. or  i  a  peck  =  1  dry  gal.  =  2G8.8  cu.  in.  nearly. 

2.  1  qr.  =  8  bu.  =  480  Ib.,  used  in  England  in  measurins  wheat. 
3G  bu.,  and  in  some  places  32  bu.  make  1  chaldron,  used  in  some 
of  the  United  States,  and  formerly  in  Great  Britain,  in  measuring 
coal ;  but  now  coal  is  bought  and  sold  by  weight  in  England,  and 
in  many  parts  of  this  country. 

3.  Grain  is  often  bought  and  sold  by  weight.     In  England,  and 
in  many  of  the  United  States,  GO  Ib.  of  wheat,  48  Ib.  of  barley, 
56  Ib.  of  rye  or  corn,  and  32  Ib.  of  oats,  are  each  declared  to  make 
i  bushel. 

ART.  193.  The  unit  of  our  dry  measure  is  the  Win- 
chester bushel,  formerly  used  in  England,  and  so  called 
from  the  town  where  the  standard  was  kept.  It  is  8  in. 
deep,  and  18A  in.  diameter,  and  contains  2150.42  cu.  in., 
or  77. 627413  Ib.  av.  of  distilled  water  at  39°  .83  Fah- 
renheit, the  barometer  at  30  inches. 

The  New  York  bushel  is  equal  to  the  imperial  bushel  of  Grea» 
Britain,  and  therefore  contains  2218.192  cu.  in. 

COMPARISON  OF    MEASURES. 

ART.  194.  The  wine  gallon  contains  231  cu.  in.;  the 
beer  gallon  282  cu.  in.;  and  the  dry  gallon  268.8  cu.  in. 

REVIEW.— ISO.  What  is  said  of  tho  bbl.  and  hhd.  ?  190.  What  is  our 
standard  of  liquid  measure?  What  does  it  contain?  191.  What  is  beer 
measure?  Repent  the  table.  What  docs  the  beer  gallon  contain  ?  192.  What 
is  ilry  measure?  Repeat  tho  table.  How  much  is  a  qr.  of  wheat?  How 
is  grain  measured?  193.  What  is  tho  unit  of  dry  mousuro?  What  doei 
it  contain  T  What  is  tho  New  York  bushel  ? 


COMPOUND  NUMBERS.  139 

These  wcro  superseded  in  Great  Britain,  in  1826,  bj 
the  imperial  gallon,  both  of  dry  and  liquid  measure, 
which  contains  277. 274  cu.  in.,  or  10  Ib.  av.  of  distilled 
water  at  G2°  Fahrenheit,  the  barometer  at  30  inches. 
At  the  same  time  the  dry  or  Winchester  bushel  was  re- 
placed by  the  imperial  bushel  of  8  imperial  gallons,  con- 
taining 2218. 192  cu.  in. 

1  wine  gallon  of  United  States  =    231    cubic  inches. 

1  beer  gallon =    282          do. 

1  dry  gallon =    208.8       do. 

1  imperial  gallon  of  G.  Britain 

for  dry  and  liquid  measure  =    277.274  do. 

1  dry  bushel  of  U.  S =  2150.42    do. 

1  imperial  bushel  of  G.  Britain  =2218.192  do. 

This  table  is  useful  in  converting  denominations  of  one  mcasun 
to  those  of  another. 

APOTHECARIES'   FLUID   MEASURE 

ART.  195.  Is  used  for  measuring  all  liquids  that  enter 
into  the  composition  of  medical  prescriptions. 

GO  minims  (it\) make  1  fluid  drachm,  (f3)- 

8  f3 1  fluid  ounce,  (f3). 

1G  f5 1  pint,  (0). 

8  0 1  gallon,  (Cong.) 

NOTES. — 1.  Cong,  is  an  abbreviation  for  conffiarium,  the  Latin  for 
gallon;  0.  is  the  initial  of  octam,  the  Latin  for  one-eighth,  the  pint 
being  one-eighth  of  a  gallon. 

2.  For  ordinary  purposes,  1  tea-cup  =  2  wine-glasses  =  8  table- 
spoons =  32  tea-spoons  =  4  f  5. 

ART.  196.  TIME. 

GO  seconds  (sec.) make  1  minute,  (min.) 

GO  min 1  hour,  (  hr.) 

24  hr 1  duy,  (da.) 

7  da, 1  week,  (  wk.) 

4  wk 1   month,  (mon.) 

12  calendar  mon 1  year,  (yr.) 

3G5  da 1  common  year. 

300  da 1   leap  year. 

100  yr 1  century,  (cen.) 

NOTE— 1  Solar  year-=3G5  da.  5  hr.  48  min.  48  sec.  =  365  J  d» 
oearly. 


140  RAY'S   HIGHER   ARITHMETIC. 

The  denomination  from  which  the  preceding  table  is 
constructed,  is  the  day,  which  is  the  interval  of  time  bo 
tween  one  mean  noon  and  the  next. 

The  apparent  noon  is  the  moment  when  the  sun  comes  to  the 
meridian  of  any  place,  and  appears  exactly  half-way  between  rising 
and  setting.  Owing  to  the  unequal  motion  of  the  earth  around  the 
eun,  and  the  oblique  position  of  its  axis  to  its  orbit,  the  interval 
between  any  apparent  noon  and  the  next  is  not  uniform. 

The  average,  however,  is  taken  of  all  these  intervals  that  occur  in 
a  year,  and  this  average  interval  is  called  the  mean  solar  day, 
which  is  divided  as  above. 

A  year  is  the  time  during  which  the  earth  makes  a  com- 
plete circuit  about  the  sun,  and  reaches  again  a  given  point 
in  its  orbit;  it  contains  365  da.  5  hr.  48  rnin.  48  sec.,  or 
nearly  3651  days. 

The  ancients  were  unable  to  find  accurately  the  number  of  days 
in  a  year.  They  had  10,  afterward  12  calendar  months,  correspond- 
ing to  the  revolutions  of  the  moon  around  the  earth.  In  the  time 
of  Julius  Ccesar  the  year  contained  365|  clays;  instead  of  taking 
account  of  the  |  of  a  day  every  year,  the  common  or  civil  year  was 
reckoned  365  days,  and  every  4th  year  a  day  was  inserted,  ( called 
the  intercalary  day,)  making  the  year  then  have  36G  days.  The  extra 
day  was  introduced  by  repeating  the  24th  of  February,  which  with 
the  Romans  was  called  the  sixth  day  before  the  kalends  of  Jfarch.  The 
years  containing  this  day  twice,  were  on  this  account  called  bissex- 
tile, which  means  having  two  sixths.  By  us  they  are  generally  called 
leap  years. 

But  365]  days,  -—  365  days  and  6  hours,  is  a  little  longer  than  the 
true  year,  which  is  365  days  5  hours  48  minutes  48  seconds.  The 
difference,  11  minutes  12  seconds,  though  small,  produced,  in  a  long 
course  of  years,  a  sensible  error,  which  was  corrected  by  Gregory 
XIII.,  who,  in  1582,  suppressed  the  10  days  that  had  been  gained,  by 
decreeing  that  the  5th  of  October  should  be  the  15th. 

To  prevent  difficulty  in  future,  it  has  been  decided  to 
adopt  the  following  rule. 

REVIEW. — 194.  Repeat  the  table  for  comparison  of  measures.  195.  AVhai 
is  apothecaries'  fluid  measure?  Repeat  tho  table.  196.  Repeat  the  table 
of  time.  What  is  tho  unit  of  this  table?  What  is  a  mean  solar  clay? 
A  year?  What  did  the  mouths  correspond  to?  Wbnt  was  the  Julian  cal- 
endar? How  many  days  were  taken  for  a  year  in  it?  Whnt  i?  tiie  true 
length  of  tho  solar  year?  What  error  was  committed  in  tho  Julian  calen- 
dar ?  When  and  by  \rbom  was  it  corrected  ?  What  is  n,  leap  year  ? 


COMPOUND   NUMBERS.  HI 

RULE  FOR  LEAP  YEARS. 

Every  year  that  is  divisible  by  4  is  a  leap  year,  unless  it 
ends  with  a  double  cipher;  in  which  case  it  must  be  divisible  by 
400  to  be  a  leap  year. 

Thus,  1832,  1648,  1600  and  2000  are  leap  years;  but 
1857,  1700,  1800,  1918,  are  not. 

fhe  Gregorian  calendar  was  adopted  in  England  in  1752.  The 
rrror  then  being  11  days,  Parliament  declared  the  3d  of  September 
w>  be  the  14th,  and  at  the  same  time  made  the  year  begin  January 
1st,  instead  of  March  25th.  Russia,  and  all  other  countries  of  the 
Greek  Church,  still  use  the  Julian  calendar;  consequently  their 
dates  (Old  Style)  are  now  12  days  later  than  ours,  (New  Style). 
The  error  in  the  Gregorian  calendar  is  small,  amounting  to  a  day 
in  3GOO  years. 

ART.  197.  The  names  of  the  months  in  their  order, 
are  January,  February,  March,  April,  May,  June,  July, 
August,  September,  October,  November,  December. 

The  year  formerly  began  with  March  instead  of  Janu- 
ary; consequently,  September,  October,  November  and 
December  were  the  7th,  8th,  9th,  and  10th  months,  as 
their  names  indicate;  being  derived  from  the  Latin  nu- 
merals Septem  (7),  Octo  (8),  Novem  (9),  Decem  (10). 

ART.  198.   MISCELLANEOUS  TABLE. 

12  things make  1   dozen. 

12  dozen  or  144  things 1  gross. 

12  gross  or  144  dozen 1  great  gross. 

20  things 1  score. 

56  Ib 1  firkin  of  butter. 

100  Ib 1  quintal  offish. 

19G  Ib 1  bbL  of  flour. 

200  Ib 1  bbl.  of  pork. 

14  Ib 1  stone. 

2lA  stone 1  pig  cf  iron  or  lead. 

8  pigs 1  father. 

ART.  199.    The   words   folio,  quarto,  octavo,  &c.,  used 

REVIEW. — 196.  What  is  the  rule  for  leap  years  in  the  Gregorian 
calendar?  AVhero  docs  the  Julian  calendar  still  prevail?  What  is  tha 
error  in  the  Gregorian  calendar  ?  197.  Name  the  months  in  their  order  I 
What  is  the  origin  of  the  names,  September,  October,  <tc.? 


142  RAY'S   HIGHER   ARITHMETIC. 

in   speaking   of  books,    show   how   many   leaves   make  a 
Bbcet  of  paper. 

A  sheet  folded  into  2  leaves  forms  &  folio size. 

Do 4. ..do quarto  or  4to do. 

•Do 8. ..do octavo  or  8vo do. 

Do 12. ..do duodecimo  or  12nu>.   do. 

Do 18.. .do 18mo do. 

Do 30. ..do 3Gmo do. 

Also, 

24  sheets  of  paper  make 1  quire. 

20  quire? 1  ream. 

2  reams 1   bundle. 

5  bundles 1  bale. 

CIRCULAR   OR  ANGULAR  MEASURE 

ART.  200.  Is  used  for  latitude  and  longitude,  and  for 
expressing  the  distances  between  any  two  points  on  the 
surface  of  the  globe,  or  in  the  heavens. 

00    seconds  (")  make  1  minute,  (') 

GO' 1  degree,  (°) 

1*0° 1  sign,  (*.) 

300°  or  12s 1  circumference,  (c.) 

NOTE. — 'jO°=  1  quadrant  or  quarter  of  a  circumference,  180°  =>• 
1  semi-circumference.  The  degree  being  3 ,'  ^  of  a  circumference  is 
of  different  lengths  in  different  circumferences;  thus,  the  equator 
being  larger  than  the  polar  circles,  a  degree  of  the  former  is  larger 
than  a  degree  of  the  latter. 

COMPARISON    OF   TIME   AND   LONGITUDE. 

ART.  201.  The  difference  of  longitude  of  two  places  is 
the  distance  in  degrees,  minutes,  and  seconds,  that  one  of 
them  is  further  cast  or  west  of  the  established  meridian 
than  the  other. 

The  sun  appears  to  go  entirely  round  the  earth,  (3GO°), 
from  cast  to  west,  in  24  hours,  crossing  in  succession  tbe 
meridians  of  all  places  on  its  surface. 

REVIEW. — 200.  What  is  circular  or  angular  measure?  Repeat  the 
table.  201.  AVhat  is  the  difference  of  longitude  of  two  places?  At  what 
rate  per  hour  does  the  sun  appear  to  travel  around  the  earth  ?  At  what 
rato  per  minute-  ?  What  per  second  7 


COMPOUND  NUMBERS.  143 


A  place  farther  east  than  another  will  have  the  sun  on 
its  meridian  sooner,  and,  therefore,  its  time  will  be  later 
at  the  rate  of  1  hour  for  every  15°  of  longitude;  or,  (by 
taking  c'0  of  each  of  these  quantities),  at  the  rate  of  1 
minute  for  every  15'  of  longitude;  or  (by  taking  s'0  of 
these),  at  the  rate  of  1  second  for  every  15"  of  longitude 
JJence. 

1  hour  of  time =  15°  of  longitude. 

1  minute    do   =  15'         do. 

1  second    do   =  15"        do. 

NOTE. — Recollect  that  if  one  place  hat  greater  east  or  less  west  lon- 
gitude than  another,  its  time  must  be  later ;  and  conversely,  if  one  place 
has  later  time  titan  another,  it  must  have  greater  east  or  less  west  longitude. 

MONEY  TABLES. 

FEDERAL  OR  UNITED  STATES  MONEY 

ART.  202.    Is  the  currency  of  the  United  States. 

10  mills  (m.) , make  1  cent,  (ct.) 

10  ct 1  dime,  (d.) 

10  d 1  dollar,  ($). 

$10  1  eaglo. 

NOTE. — The  cent  and  mill,  which  are  yjg  and  jo'jjTJ  °f  a  dollar, 
derive  their  names  from  the  Latin  centum  and  mille,  meaning  a  hun- 
dred, and  a  thousand;  the  dime  which  is  7*5  of  a  dollar,  is  from  the 
French  word  disme,  meaning  ten. 

ART.  203.  The  Federal  currency  was  authorized  by  act 
of  Congress,  August  8th,  1780.  It  has  great  simplicity, 
being  on  the  decimal  basis,  and  subject  to  the  law  that 
one  of  any  denomination  is  equal  to  10  of  the  next  lower; 
therefore,  the  same.  Notation,  Numeration,  and  general 
order  of  operation,  can  be  used  for  Federal  money  as  for 
simple  numbers. 

Any  sum  of  Federal  money  of  several  denominations  can  be  ex- 
pressed as  one  denomination,  by  writing  those  of  that  denomination 
in  units'  place,  those  of  higher  denominations  in  places  of  whole 
numbers,  and  those  of  lower  denominations  in  decimal  places ;  thus, 

REVIEW. — 201.  Why  will  a  pines  that  is  east  of  another  have  later 
time?  What  is  tho  table  for  comparing  time  and  longitude?  How  do  w« 
know  from  tho  longitudes  of  two  places,  which  hag  later  time?  202.  \Vhal 
u  Federal  or  U.  S.  money?  Repeat  the  table. 


.  144  RAY'S   HIGHER   ARITHMETIC. 

>  eagles  $7  2  Duties  4  ct.  and  6  m.  can  be  written  9.7245  eagki», 
in  .s'lT.'Jlo,  or  972.45  dimes,  or  9724.5  ct.  or  97245  m.  It  is  cus- 
tomary to  consider  the  dollar  as  the  unit,  and  express  all  sums  of 
U  S.  money  in  that  denomination  and  its  decimal  divisions. 

In  reading  U.  S.  money,  name  the  dollars  and  all  higher 
denominations  together  as  dollars,  the  dimes  and  cents  at 
cents,  and  the  next  figure,  if  there  is  one,  as  mills; 

Or,  name  the  whole  numbers  as  dollars,  and  the  rest  as 
a  decimal  of  a  dollar. 

Thus,  $9.124  is  read  9  dollars  12  ct.  4  mills,  or  9  dollars  124  thou- 
sandths of  a  dollar;  $175.06238  is  read  175  dollars  Get.  2  mills, 
and  a  remainder,  or,  175  dollars,  6238  hundred-thousandths  of  a 
dollar. 

ART.  204.  The  national  coins  of  the  United  States  are 
of  gold,  silver,  and  copper.  The  gold  coins  are  the  double- 
eagle,  eagle,  half-eagle,  quarter-eagle,  and  one  dollar  piece. 
A  3  dollar  piece  has  also  been  authorized. 

The  silver  coins  are  the  dollar,  half-dollar,  quarter-dollar, 
dime,  half-dime,  and  3  cent  piece. 

The  copper  coins  are  the  cent  and  half-cent;  the  latter 
is  now  obsolete. 

The  mill  never  has  been  a  coin;  it  is  only  a  convenient  name  for 
the  tenth  of  a  cent,  or  thousandth  of  a  dollar. 

Pure  gold  and  silver  being  too  soft  for  coins,  are  mixed 
with  baser  metal,  called  aUoy.  By  act  of  Congress,  in 
1837,  our  standard  gold  and  silver  is  7%  pure  and  -fa  alloy, 
(by  weight). 

The  alloy  in  the  silver  coins  is  pure  copper;  in  the  gold 
coins  it  is  copper  and  silver,  the  latter  not  to  exceed  the 
former  in  weight. 

The  3  cent  piece  is  not  standard  silver,  being  one-fourth  copper. 
It  weighs  12|  grains.  The  cent  weighs  168  gr. 

REVIEW.— 202.  Why  is  the  cent  so  called?  The  mill?  The  dime? 
203.  How  do  the  denominations  of  Federal  money  resemble  those  of  simple 
numbers  ?  How  can  sums  of  Federal  money  be  read,  written,  added,  sub- 
tracted, <fec.?  In  writing  any  sum  of  Federal  money,  what  single  denomi- 
nation is  generally  used  ?  In  reading  any  sum  of  Federal  money,  which 
denominations  only  nro  mentioned?  What  single  denomination  may  be 
employed?  204.  Which  coins  are  of  gold?  Which  of  silver?  Which  of 
copper?  What  is  the  mill?  What  are  standard  gold  and  silver?  What 
is  the  alloy  for  gold?  What  for  silvsr? 


COMPOUND  NUMBERS.  145 


The  eagle  weighs  258  gr.  The  half-dollar  since  April  1st,  1868, 
weighs  1(J'J  gr. 

The  half  dollar  coined  before  April  1st,  1853,  vreisrhs  206  {  gr.;  it 
contains  more  silver  than  the  one  now  coined,  and  is  worth  63j92g, 
instead  of  50  cents. 

ART.  205.  The  fineness  of -manufactured  gold  is  esti- 
mated in  carats,  an  Arabic  or  Abyssinian  word,  signifying 
a  small  weight.  In  determining  the"  purity  of  gold  by 
analysis,  a  portion  of  it  is  taken,  and,  without  reference  to 
its  actual  weight,  is  called  the  assay  pound,  which  is  thus 
divided : 

4  quarters  (qr.) make  1  assay  grain,  (gr.) 

4  gr 1  carat,  (car.) 

21  car 1   assay  pound 

NOTES. — 1.  Each  carat  is  ^  of  the  mass  used;  if  it  is  18  carats 
gold,  it  is  Af  pure  gold,  and  26f  alloy;  if  it  is  '2'2  carats  gold,  it 
is  I  ~\  pure  gold,  and  ^  alloy;  if  it  is  24  carats  gold,  it  is  entirely 
free  from  alloy. 

2.  The  quarters  arc  written  as  fourths  of  a  carat  grain;  thus, 
19  car.  3:[  gr. 

ENGLISH  OR  STERLING  MONEY 
ART.  206.    Is  the  currency  of  Great  Britain. 

4  farthings  (qr.)  make  1   penny,  (d.) 

1*2  ponce 1   ehilling,  (s.) 

20  shillings ..   1  pound,  (£)  =  $4.84* 

NOTES. — 1.  The  Guinea,  (gold),  =  2ls.;  crown,  (silver),  =  6s. 
half-crown  =  2s.  Gd.;  noble  =  6s.  8d. ;  angel  =  10s.;  mark  = 
13s.  4d. ;  pistole  =  lt>s.  10d.;  moidore  =  27s. ;  1  sovereign  =  20s., 
(gold)  =  $4.8-1,  «by  act  of  Congress,  1842. 

2.  The  furthing  is  not  a  coin,  but  stands  for  a  quarter  of  a 
penny;  thus,  6^d.  =  5  pence  3  farthings.  When  the  penny  was 
of  silver,  it  was  usual  to  mark  it  with  a  cross  so  deep  that  it  could 
Ij  eisily  broken  into  halves  and  quarters,  called  half-pennies  and 
ftur things,  finally,  farthings. 


.-  201.  What  is  tho  weight  of  the  eagle?  The  present  half- 
dollnr  ?  What  was  the  weight  of  the  hnlf  dollar  Wforo  1853?  How  much 
U  it  now  worth?  205.  How  is  the  fineness  of  manufactured  gold  esti- 
mated? Repent  the  tiible.  208.  What  i.«  sterling  money?  Kc|>fat  th« 
tatite.  How  are  farthing*  written  ?  Which  dououiiuutions  arc  not  coiz>*  7 
What  13  the  origin  of  the  nimo /artking  t 
13 


146  RAY'S   HIGHER   ARITHMETIC. 


3.  The  wind  (£.}  is  not  a  coin,  but  stands  for  20s.;  it  is  repre- 
Bented  oy  i-be  sovereign,  or  the  bank  note  of  £1.  The  pound  is  BO 
called,  because  its  equivalent,  240d.  or  20s.,  formerly  contained  a 
pound  weight  of  silver,  the  pound  then  being  smaller  than  at 
present.  A  pound  of  standard  silver  is  now  coined  into  66s. 

REMARKS. — 1.  The  symbols,  £.,  s.,  d.,  q.,  are  the  initials  of  the 
Latin  words  libra,  solidarius,  denarius,  quadrans;  signifying,  respect- 
vely,  pound,  shilling,  pcnnv,  and  quarter. 

2.  The  sovereign,  the  standard  gold  coin,  weighs  123.274  gr., 
and  is  22  carats  fine.  The  shilling,  the  standard  silver  coin,  is  |g 
pure  silver,  and  fa  copper,  and  weighs  87.27  gr.  Pence  and  half- 
pence are  made  only  of  copper  now,  and  each  penny  weighs  240 
gr.  =  2  oz.  Troy. 

OF  STATE   CURRENCIES. 

ART.  207.  Before  our  present  currency  was  established, 
our  accounts  were  kept  in  pounds,  shillings,  and  pence. 

In  many  States,  the  denominations  shillings  and  pence 
are  still  retained,  but  not  with  the  same  values. 

In  New  Hampshire,  Massachusetts,  Rhode  Island,  Connecticut, 
Virginia,  Kentucky,  and  Tennessee, 

12d.=    1  shilling^  IGf  cents. 
6s.  =$1  =  100  cents. 

In  New  York,  North  Carolina,  and  Ohio, 

12d.=    1  shilling  =12^  cents. 
8s.  =$1  =  100  cents. 

In  New  Jersey,  Pennsylvania,  Delaware,  and  Maryland, 

12  d.  =    1  shilling  =13£  cents. 
7s.  Gd.  or  9Cd.  =  $1  =  100  cents. 

Ic  South  Carolina  and  Georgia, 

12d.  =  1  shilling  =  21 2  cents. 
4s.  8 d.  or  56d.  =  $1  =  100  cents. 

la  Canada  and  Nova  Scotia, 

12d.=     1  shilling  =  20  cents. 
5s.  =  $1  =  100  cents. 


REVIEW. — 206.  What  is  tho  origin  of  tho  name  pound?  Into  how 
many  shillings  \»  a  pound  of  silver  now  coined?  207.  Whnt  is  tho  Now 
England  State  currency  ?  NawYork?  Pennsylvania?  Georgia?  Canada* 


COMPOUND  NUMBERS. 


FRENCH  WEIGHTS  AND  MEASURES. 

ART.  208.  The  present  system  of  weights  and  measures 
in  France,  adopted  in  1795,  is  on  the  decimal  basis. 

After  the  unit  of  any  measure  has  been  determined  and 
named,  the  higher  denominations  are  made  by  prefixing 
the  Greek  numerals,  deca  (10),  heclo  (100),  kilo  (1000), 
myrin  (10000)  to  the  name  of  the  unit;  the  lower  de- 
nominations are  formed  by  prefixing  the  Latin  numtrali 
deci  (jo),  centl  (TOO),  milli  (TO'OO). 

FRENCH   LONG   MEASURE. 

ART.  209.  The  unit  of  long  measure  in  France  is  the 
mitre,  which  is  the  ten-millionth  part  of  the  quadrant,  ex- 
tending through  Paris  from  the  equator  to  the  pole. 

1  metre =39.371  U.  S.  in. 

NOTE. — 1  decametre  =  10  metres;  1  hectometre  —  100  metres; 
1  kilometre  =  1000  metres;  1  my  riametre  =  10000  metres;  1 
decimetre  =  y'g  metre;  1  centimetre  =  y Jo  metre;  1  millimetre 
=  TO'OO  metre. 

FRENCH  SURFACE  MEASURE. 

^XART.  210.    The  unit  of  surface   in   France  is  the  are, 
which  is  a  square  decam6tre. 

1  are =  119.6046  U.  S.  sq.  yd. 

NOTE. — 1  decare  =  10  ares ;  1  hectare  =  100  ares ;  1  ccntiare 
=  TOO  are. 

FRENCH   SOLID    MEASURE. 

ART.  211.  The  unit  of  solidity  in  France  is  the  sttre, 
which  is  a  cubic  metre. 

1  store =35.31741  U.  S.  cu.  ft. 

NOTE. —  1  decastere  =  10  stores.     1  decistere  =  y'g  stere. 

FRENCH    WEIGHTS. 

ART.  212.  The  unit  of  weight  in  France  is  the  gramme, 
which  is  the  weight  of  a  cubic  centim6tre  of  distilled  water 
at  the  temperature  of  melting  ice. 

1  gramme =  15.434  Troy  gr. 

NOTI. — 1  decagramme  =  10  grammes;  1  hectogramme  =  100 
Rrnuiniea;  1  kilogramme  =  1000  grammes  -  2^  Ih.  av.  nearly; 


14 a  HAY'S   HIGHER   ARITHMETIC. 

1  niyri:igra;,ime  =  10000  grammes.  1  decigramme  =  j'0  grnmroc; 
1  centigramme  ••  j,1^  gramme;  1  milligramme  =  75*00  gramme; 
1  quiutal  =  220.55  Ib.  av.;  1  millier  or  bar  =  2205.5  Ib.  av. 

FRENCH   DRY   AND   LIQUID    MEASURE. 

AR?  213.  The  unit  of  capacity  in  France  is  the  litre, 
*hioh  is  a  cubic  decimetre. 

1  litre     ....  =2.1135  pt.  wine  measure,  U.  S. 

NOTE. — 1  decalitre  =  10  litres;  1  hectolitre  =  100  litres;  1 
kilolitre  =  1000  litres  ;  1  myrialitre  =  10000  litres.  1  decilitre  = 
,-'0  litre;  1  centilitre  =  700  litrci  1  millilitre  =  70*00  litre. 

ART.  214.  Soms  of  the  old  weights  and  measures  arc 
used;  as,  1  livrc  =  A  a  kilogramme;  1  marc  =  i  a  livre; 
1  once  =  A  mare;  1  gros  =  g  once;  1  grain  =  ^ 
pros:  1  toise  =  2  metres;  1  pied  or  foot  =  ^  metre; 
1  inch  =  y'r,  pied  or  foot ;  1  aunc  =  11  metres;  1  bois- 
BCau  or  bushel  =  12  j  litres;  1  litron  =  1.074  Paris 
pints.  When  these  are  employed,  the  word  mud  is  an- 
nexed to  them,  signifying  customary. 

FRENCH    MONEY. 

ART.  215.  The  unit  of  money  is  the  franc,  which  is  79u 
pure  silver,  and  7*0  alloy,  like  our  silver  coins. 

1  franc =$.1875  or  18.f  cents. 

NOTE. — 1  decime  =  7*0  franc;     1  centime  =  7^5  franc. 

The  livre  lournois,  the  former  unit  of  French  money,  =  18\  cents. 

FRENCH   CIRCULAR   OR  ANGULAR   MEASURE 

ART.  216.  Is  the  same  as  that  of  the  United  States 
and  other  countries. 

FOREIGN  WEIGHTS  AND    MEASURES. 

ART.  217.  The  pounds  here  mentioned  are  Ib.  avoir- 
iupcis,  when  not  otherwise  specified. 

dlcjcnnrlria,  Egypt. — 1  pik  =  20.8  in.;  1  rhebebe  =  4.3G4  bu. ; 
1  'inillot  or  kisloz  =  4.729  bu.:  1  rcltolo  for foro  =  .9347  11). 
ar.;  1  rottolo  zaidino  =  1.335  Ib.  av. ;  1  rottolo  zaro  =  2.07 
Ib.  av. ;  1  rottolo  mina  =  1.G7  Ib.  av. ;  1  quintal  or  cuntaro 
=•  100  rottoli. 


FOREIGN   WEIGHTS   AND   MEASURES.  14<, 

Amsterdam,  Holland. — French  system  adopted  in  1820.  01(1 
measures  as  follows:  1  lb.  =1.08923  lb. ;  1  last  =  85.25  bu. ; 
1  aarn  =  41  wine  gal.;  1  stoop  =  5^  pints  ;  1  anker  =  I0j 
gal. ;  1  foot  =  11}  in. ;  1  ell  =  27  ,-*  in. 

Antwerp,  Belyium. — French  system  adopted  in  181G.  Old  mea- 
sures as  follows:  1  lb.  =  1.03$  lb.  av. ;  1  suhippound  ~  & 
quintals  =  310  lb.  av. ;  1  aam  =  3GA  w.  gal.;  1  vicrtel  -•=» 
2.125bu.;  1  stoop  =  5.84  pt. 

Barcelona,  North  of  Spain. — 1  lb.  =  .88215  lb.  av. ;  1  cana  =• 
.58514  yd. ;  1  quartern  =  1.^8288  bu. ;  1  carga  =  32.7  w.  gal 

Bombtty,  East  Indies. — 1  maund  =  28  lb.  av. ;  1  candy  =  560 
lb.  av. ;  1  tola  =  179  Troy  gr. ;  1  tank  for  pearls  =  72  gr.  • 
1  guz  =  27  in.  ;  1  hath  =  18  in.;  1  tursoo  =  1^  in. 

Bremen.— I  lb.  =  1.098  lb.  av. ;  1  last  —  80.7  bu. ;  1  aam  or  4 
sinkers  =  37|  w.  gal. ;  1  foot  =  11.38  in. ;  1  ell  =  .032  yd. 

Batacia,  East  Indies. — 1  pecul  =  136  lb.  av. ;  1  catty  =  1.36 
lb.  av. 

Bcncookn,  East  Indies. — 1  bahar  —  5GO  lb.  av. 

Baltia  and  liio  de  Janeiro,  Brazil. — I  alquicrc  of  grain  =  1  bu., 
U.  S. ;  1  frasco  =  4.5  pt.  • 

Cadiz,  South  of  Spain.— I  lb.  =  1.015  lb.  av. ;  1  vara  =  .9275 
yd. ;  1  arroba  of  wine  =  4{  w.  gal. ;  1  arroba  of  oil  =  3  4  w. 
gal.;  1  mo  jo  =  68  w.  gal. ;  1  botta  =  127  A  w.  gal. 

Calcutta  and  Bengal  Factory,  East  Indies. — 1  maund  —  74j  lb. 
av. ;  1  bazaar  maund  =  82fj  lb.  ;  1  tolan  =  224.588  T.  gr., 
1  pallic  =  9.08  lb.  av. ;  1  chittack  =  45  sq.  ft. ;  1  biggah  = 
14440  sq.  ft.  ;  1  guz  —  1  yd.;  1  coss  =  I.**  miles. 

Canton,  China. — 1  tael  =  1|  oz.  ;  1  catty  =  lj  lb.  av. ;  1  pecul 
=  133]  lb.  av.  ;  1  covid  or  cobre  =  14.025  in. ;  1  li  = 
1«97.1,  ft. 

Con.ttnntinople,  Turkey. — 1  quintal  or  cantaro  =  124.457  lb.  av. ; 
1  quintal  of  cotton  =  127.2  lb.  av. ;  1  pik  of  silk  =  27.9  in.  j 
1  pik  of  cotton  =  27  in. ;  1  kisloz  =  .741  bu. ;  1  alma  of  oil 
=  U  gal. 

Copenhagen,  Denmark. — 1  lb.  =  1.1025  lb.  av. ;  1  anker  =  10  w 
gal.;  lpot=1.02qt;  1  last  =  380  bu. ;  1  Rhineland  fool 
=  12 «  in.;  1  Danish  ell  =  2.06  ft. 

Datitzic,  East  Prussia.—  1  lb.  =  1.033  lb.  av.;  1  last  =  020.4  w. 
gal.;  1  alim  of  wine  -=  39^  w.  gal.;  1  sclicflfel  =  1.552  bu. ; 
1  D:mtxic  foot  =  11.3  in.;  1  Uhinoland  or  1'russian  foot  = 
12.356  in.;  1  Prussian  ell  =  26.250  in.;  1  last  of  corn  =  91 
bu. ;  1  last  of  wheat,  rye  =  87  bu.;  1  Pru.  mile  —  4.8  milea 


|f»(>  RATS   ILGIIER   ARITHMETIC. 


G'cnna,  Sardinia. —  1  Ib.  peso  sottile,  =  .0989  Ib.  av. ;  1  Ib., 
grosso,  =  .70875  Ib.  av. ;  1  mina  =  S.j  bu. ;  1  mezzarola — 
39.}  w.  gal  ;  1  barilla  of  oil  =  17  w.  gal.;  1  palmo  =  9.72v 
in.;  1  canna  =  9,  12  or  10  palmi,  as  it  is  used  by  manufac- 
turers, merchants,  or  custom-house  officers. 

Hamburgh.— \  Ib.  =  1.008  Ib.  av. ;  1  ahm  =  38y  w.  gal. ;  1 
fuder  =  229j  Sal- ;  1  steckan  of  oil  =  5  3  gal. ;  1  last  =  89.6 
bu.:  1  foot  =  11.289  in. ;  1  Brabant  ell  =  27.585  in. 

Havana,  Cuba. — 1  arroba  of  wine  or  spirits  =  4.1  w.  gal.;  1 
funega  =  3  bu.,  nearly;  1  arroba  of  weight  or  25  Ib.  =  25.4375 
Ib.  av. ;  1  vara  =  2$  ft. 

Konigsberg. — Same  as  Dantzic. 

La    Guayra,  Venezuela. — Same  as  Spain. 

Lei/horn,  Tuscany. — 1  Ib.  =  .74804  Ib.  av.,  generally  reckoned,  — 
.77  Ib.  av. ;  1  sacco  of  corn  =  2.0739  bu. ;  1  barile  =  12  w. 
gal. ;  1  braocio  =  22.98  or  23  in. ;  1  canna  =  92  in. 

Lima,  Peru. — Same  as  Spain. 

Lisbon,  Portugal. — 1  Ib.  or  arratel  =  1.10119  Ib.  av. ;  1  moyo  = 
23.03  bu. ;  1  almude  =  437  w.  gal.;  1  tonelada  =  227 \  w. 
gal. ;  1  pipe  of  Lisbon  =  140  w.  gal. ;  1  pipe  of  port  =  108  w. 
gal. ;  1  pe  or  foot  =  12.944  in. ;  1  vara^=  43.2  in. ;  1  Al- 
mude of  Oporto  =  Gg  w.  gal. 

Madras,  East  Indies. — 1  maund  =  25  Ib. ;  1  candy  =  500  Ib. ; 
1  garce  =  137  bu. ;  1  Company  maud  =  24|  Ib. ;  1  varahun 
=  52?  gr. ;  1  visay  =  3  Ib.  3  dr. ;  1  baruay  =  482.}  Ib. ;  1 
gursay  =  9045  .j  Ib. 

Montevideo,  liuenos  Ayres. — Same  as  Spain. 

Muscat,  Arabia. — 1  maund  or  24  cuchas  =  8?  Ib.  av. 

Naples,  Naples. — 1  cantaro  grosso  =  190  \  Ib.  av. ;  1  cantaro 
piccolo  =  100  Ib. ;  1  tomclo  =  1.45  bu. ;  1  carro  =  264  w. 
gal.  ;  1  pipe  wine  or  ludy  =  132  w.  gal.  ;  1  salma  =  423 
w.  gal. ;  1  canna  =  6  ft.  11  in. ;  1  palmo  =  10.375  in. 

Odessa,  Russia. — Same  as  St.  Petersburg. 

Palermo,  Sicily. — 1  oncie  =  {  5  oz. ;  1  salma  grossa  =  9.48  bu.  ; 
1  salma  generale  =  7.02  bu. ;  1  barile  =  9^  w.  gal.;  1  caffiso 
of  oil  =  4 if  w.  gal. ;  1  canna  =  3.',  yd. ;  1  palma  =  1  i  ft. 

Port-au-Prince,  Ifayti. — Measures  same  as  France — weights  same 
as  England,  but  8  per  cent,  heavier. 

Purld- It ico. — Same  as  Havana. 

\an;/fif>n,  Kant  Indies. — 1  kyat  or  tical  =.584  Ib.  av. ;  1  Paiktha  or 
vis  =  3. 05  Ib.  av. ;  1  ten  or  basket  =  58.4  Ib.  av.,  generally 
reckoned  .1  owL 


REDUCTION   OF   COMPOUND   NUMBERS.  151 

Ri'ja,  Russia.— I  Ib.  =  .9217  Ib.  av. ;  1  loof  =  1.9375  bu. ;  1 
anker  =  10^  w.  gal.  ;  1  foot  =  10.79  in.,  U.  S. 

Rotterdam,  Holland. — 1  last  =  10.542  bu.  ;  1  ahm  =  40  w.  gal., 
nearly;  1  stoop  =  .6775  w.  gal ;  1  foot  =  1.02  ft,  U.  S. 
The  rest  like  Amsterdam. 

Singapore,  East  Indies. — 1  maund  of  rice  =  82. 125  Ib.  av.  ;  1 
bungkal  of  gold-dust  =  832  gr.  The  rest  same  as  Canton. 

Smyrna,  Turkey. — Same  as  Constantinople.  1  rottolo  =  1.2748 
Ib.  av. ;  1  oke  =  2  Ib.  13  oz.  5  dr. ;  1  tepper  of  silk  =  4j  Ib., 
av. ;  1  chequee  of  opium  =  lg  Ib.  av. ;  1  chequee  of  goats' 
wool  =  5 1  Ib. ;  1  kellow  =  1.456  bu.  ;  1  pic  =  27  in.,  U.  S. 

Stockholm,  Sweden. — 1  Ib.  or  pund  =  .9375  Ib.,  av. ;  1  Ib.  of 
iron  =  |  Ib.  av. ;  1  tun  =  4^  bu. ;  1  ahm  =  4iy.j  w.  gal.  ; 
1  pipe  =  124J-  w.  gal. ;  1  foot  =  11.684  in.,  U.  S. ;  1  kannor 
=  .692  w.  gal. 

St.  Petersburg,  Russia.—  1  Ib.  =  .9026  Ib.  av. ;  1  pood  =  36.1041 
Ib.  generally  reckoned  36  Ib.  av.  ;  1  wedro  =  3.14  w.  gal.  ;  1 
chetwert  =  5.952  bu. ;  1  sashen  =  7  ft. ;  1  arsheen  =  28 
in. ;  1  foot  =  1.145  ft,  U.  S. ;  1  verst  or  mile  =  5.3  fur. 

Trebisond,  Turkey. — Same  as  Constantinople. 

Triente,  Austria.— I  Ib.  =  1.236  Ib.  av.  ;  1  staro  =  2.34  bu. ;  1 
Vienna  nietzen  =  1.723  bu.  ;  1  polonick  =.861  bu. :  1  orna 
or  eitner  =  15  w.  gal. ;  1  barile  =  173  T  w.  gal. ;  1  orna  of 
oil  =  17  w.  gal. ;  1  ell  (for  woolen  goods)  =  2.27  in. ;  1  ell 
of  silk  =  25.2  in. 

Valparaiso,  Chili. — Same  as  Spain. 

Venice,  Lombardy. — I  Ib.,  peso  sottile  =  .66428  Ib.  av. ;  1  Ib., 
peso  grosso  =  1.05186  Ib.  av. ;  1  etaja  =  2.27  bu.  ;  1  an  fora 
=  137  w.  gal. ;  1  miro  =  4.028  w.  gal. ;  1  braecio  of  wool 
=  26.6  in. 

Vera  G-uz,  Mexico. — Same  as  Spain. 


REDUCTION  OF  COMPOUND  NUMBERS. 

ART.  218.    Reduction  is  changing  the  form  of  a  num- 
ber without  altering  its  value.     It  has  three  cases. 

CASE  I. — To  reduce   a  simple  number  of  any  denomi- 
nation to  another  denomination. 

GENERAL   RULE. 

Multiply  the  given  number  by  its  unit  value  in  the  Jenomin* 
Sion  required' 


152  RAY'S   HIGHER   ARITHMETIC. 

Or,  Diciile  it  by  the  unit  value  of  the  required  denomination 
in  the  one  (jiren. 

XOTK. —  When  there  are  one  or  more  denominations  between  the 
one  given  and  the  one  required,  redace  the  given  number  to  eac* 
of  ihera  in  succession,  until  the  required  denomination  is  reached. 

Reduce  18  bushels  to  pints. 

SOLUTION. — Since   1  bu.  =  4   pk.,  18  bu.  =  18  18  bu. 

imes  4  pk.  =  72  pk.,  and  since   1  pk.  =  8  qt.,  72  4 

pk.  =  72   times  8  qt.  =  570  qt.,  and   since   1  qt.=  ^  Q     , 

2  pt.,  576  qt.=  576  times  2  pt.  =  1152  pt_     Or,  find 
the  unit-value  of  bushels  in  pints,  thus:   1  bu.  =  8 
pk.  =32qt.  =  04  pt.;     then   multiply  18  by  64  pt.          576  <!&• 
which  gives  1152pt.  as  before.     This  is  called  Re-  2 

duction  Descending,  that  is,  going  from  a  hit/her  de-      115°  r>t 
nomination  (bu.)  to  a  lower  (pt.)  and  is  performed 
by  the  1st  part  of  the  rule,  that  is,  by  multiplication. 

Reduce  236  inches  to  yards. 

SOLUTION.— Since  12  inches  =  1  ft,,  236     12)23G  in. 
inches  will  be  as  many  feet  as  12  in.  is  con-      ~oT     -i  Q.J 

tained   times  in  236  in.,  which  is  10^  ft.,  and      I ll 

since  3  ft.  =  1  yd.,  10J  ft.  will  be  as  many  yd.  65  yd. 

as  3ft.  is  contained  times  in  19j  ft.  which  is 

6 1  yd.  Or,  find  the  unit-value  of  yards  in  inches,  thus;  1  yd.  =» 
3ft.  =36  in.;  then  divide  230  in.  by  30  in.,  which  gives  tijj  yd.,  aa 
before.  This  is  called  Reduction  Ascending,  that  is,  going  frorp  " 
lower  denomination  (in.)  to  a  higher  (yd.),  and  is  generally  per- 
formed by  the  2d  part  of  the  rule,  that  is,  by  division. 

ART.  219.  Reduction  Ascending  is  similar  in  principle 
to  Reduction  Descending,  and  can  be  performed  by  the  1st 
part  of  the  rule. 

Thus,  in  last  example,  instead  of  dividing  236  in.  by  36  in.  the 
unit  value  of  yards,  the  236  mny  be  multiplied  by  r?'g  yd.,  the 
unit-value  of  inches;  for,  236  X  3'$  yd.  =  V^  yd.  =  G|  )  1.  The 
tperation  by  division  is  generally  more  convenient. 

I*  EM  AH  K. — Reduction  Descending  diminishes  the  size  and,  there- 
fore, increases  the  number  of  units  given;  while  Reduction  Ascend- 

REVIEW. — 213.  What  is  reduction?  How  many  cases?  AVhat  if 
Cn*e  1st?  The  rule?  When  there  nro  ccvcml  <lcnriniin:itions  lu-rwi-cn 
the  one  given  nn.l  the  >ne  roc|iiire<l,  whnt  should  be  dune?  Exphiin  ex- 
amples 1  and  2.  What  is  Reduction  Descending?  Ilovr  is  it  generally 
performed?  AVhat  is  deduction  Aseending?  How  is  it  generally  per- 
fcrmod?  219.  How  may  it  bo  performed? 


REDUCTION   OF   COMPOUND  NUMBERS.  If. 3 

ing  increases  the  fize,  and,  therefore,  diminishes  the  numler  of  units 
given.  This  is  further  evident  from  the  fact,  tuat  uie  multipliers 
in  Reduction  Descending  are  larger  than  1;  but  in  Reduction  As- 
cending smaller  than  1. 

,  ...  SOLUTION. 

Reduce  §  gal.  to  gills.  • 

R  K  M  A  R  K  . — The  rule  also  ap-     £_X  £X$X4  =  1.-  gillg 
plies  when  the  number  to  be  re-      a 
duced   is  a   common  or  decimal 
fraction.     Indicate  the  operations  and  then  cancel. 

Reduce  5 7  gr.  to  3-  2 

SOLUTION. — Although       5          ^p       111         1 

this  is  Reduction  Ascend-     "?   8r>        7       0fl      8     H 84 

ing,  use  the  first  part  of 

the  rule,  multiplying  by  the  successive  unit  values,  J6,  \  and  |. 

Reduce  9.375  acres  to  perches.        9.375  A 

4 

37.50P  R, 
40 


Ans.  1500.p  =  1500  P. 

Reduce  2000  seconds  to  hours.  Ans.  |  hr. 

1        1       20      5 


Also,  .6428  dr.  av.  to  Ib.  av.   Ans.  .0025109375 
16).64280Q  dr. 
1C).  040175  oz. 


.0025109375  Ib. 

1.  Reduce  2}  years  to  seconds.  Ar>*.  7095GOOO  sec. 

2.  49  hours  is  what  part  of  a  week?        Am.  Jj  \vk. 

3.  Bring  1  circumference  to  seconds.  Ans.  1200000". 

4.  25"  to  the  decimal  of  a  degree.      Ans.  .00094° 

5.  17.0625  rd.  are  how  many  inches?    Ans.  3378J 

6.  Bring  4.S  ft.  to  chains,  Ans.  ^7f  chain. 

7.  Reduce  192  sq.  in.  to  sq.  yd.          Ans.  ^  sq.  yd. 

R  K  VIEW. — 219.  Docs  the  rule  npj>lv  to  fractions?  ITow  can  the  of 
tion  be  shortened  ?  In  reducing  «omtnon  or  dvciniiil  fractions,  what  rulei 
mast  be  borno  in  mind?  Ant.  The  rules  for  tho  multiplication  and  division 
of  common  and  decimal  fractions. 


154  RAY'S   HIGHER  ARITHMETIC. 

8.  63  cu.  yd.  to  cubic  inches.      Ans.  311040  cu.  in. 

9.  $117.14  to  mills.  Ans.  117140  mills. 

SUGGESTION. — In  all  reductions  of  U.  S.  money,  or  any  system 
formed  on  the  decimal  basis,  the  multiplications  or  divisions  are 
readily  performed  by  moving  the  point  to  the  right  or  left,  since  th« 
multipliers  and  divisors  are  10,  100,  1000,  &c.,  (Arts.  144  and  145). 

10.  Reduce  6.19  cents  to  dollars.  Ans.  $.0619 

11.  1600  mills  to  dollars.  Ans.  $1.60 

12.  $51  to  mills.  Ans.  '5375  mills. 
SUGGESTION . — Change  f  to  a  decimal,  then  move  the  point. 

13.  Reduce  12  lb.  av.  to  lb.  Troy.          Ans.  14T7z  lb. 
SOLUTION.— The  12      tn v  7flflfl v       1  iyc 

tl_  a        i  t  1  i.ZPN;lv'lJy/\  1  -LlO  1    /«    7 

Ib.  av.  are  first  reduced  rT        17V        — —  I^TT 

to  gr.  by  the   table  in  £/76P        12 

Art.  194,  and  the  gr.  to  ^8    12 

Tr~y  lb.  by  same  table. 

14.  Reduce  33  beer  gal.  to  wine  gal.  Ans.  40 ?  w.  gal. 

15.  36  yd,  to  ells  Flemish.  Ans.  48  E.  Fl. 
1C.    2  in.  to  qr.  Ans.  JS<{T. 

17.  .216gr.  to  oz.  Troy.  Ans.  . 00045  oz.  Tr. 

18.  £.0732  to  pence.  Ans.  17.568  d. 

19.  fib.  to  tuns.  Ans.  WooT. 

20.  47.3084sq.  mi.  to  P.  Ans.  4844380.16  P. 

21.  4-23  to  lb.  Ans.  a1*  tt>. 

22.  7  g  dr.  av.  to  lb.  Ans.  3^  lb. 

23.  50  U.  S.  bu.  to  imp.  bu.  Ans.  48.47236  nearly. 

24.  1200  inches  to  chains.  Ans.  HI  ch. 

25.  99yd.  to  furlongs.  Ana.  2°o  fur. 

26.  6. 3419  C.  to  cu.  in. 

Ans.  1402726. 8096  cu.  in. 

27.  18  fathoms  to  miles.  Ans.  Tjomi. 

28.  How   many   acres  in  a  rectangle   24 1  rd.  long  by 
16.02rd.  wide?  Ans.  2.4530625  acres. 

20.  How  many  cubic  yd.  in  a  box  6'  ft.  long  by  2j 
ft.  wide  and  3  ft.  high  ?  Ann.  1*%  cu.  yd. 

REVIEW  — 219.  How  can  reductions  in  U.  S.  money  be  performed  T 


REDUCTION    OF   COMPOUND  NUMBERS.  155 

30.  How   many  perches   in  a  rectangular  fieM  18.22 
•chains  long  by  4.76ch.  wide?  An*.  1387.6352  P. 

31.  Reduce  250  3  to  dr.  av.  Ana.  561  [75  dr.  av. 

32.  1  nautical  league  to  feet.      Ans.  18256.7088ft. 

33.  1C. 02  chains  to  miles.  Ans.  .20025  mile. 

34.  4.29  chains  to  feet.  Ana.  283.14  ft. 

35.  |  of  a  link  to  rods.  -4ns.  35  rd. 

36.  7  of  a  nail  to  ell  English.  Ana.  Vo  E.  E. 

37.  1.644  inches  to  ell  Flemish.     Ans.  .0608  E.  Fl. 

38.  35.781sq.  yd.  tosq.  in.    Ans.  46372.176  sq.  in. 

39.  256  roods  to  sq.  chains.  Ans.  640  sq.  ch. 

SUGGESTION. — First  bring  to  acres,  then  to  sq.  chains. 

40.  13]  tuns  of  round  timber  to  cords.       Ans.  4$  C. 

41.  6.15  tuns  of  hewn  timber  to  reduced  feet,  plank 
measure,  1  inch  thick.  Ans.  3600. 

42.  How  many  perches   cf  masonry  in    a  rectangular 
y    solid   wall   40ft.   long   by  7aft.  high  and   2|  ft.   average 
i      thickness?  Ans.  32 J> I  P. 

43.  Reduce  -7  *lb.  Troy  to  gr.  Ana.  1040gr. 

44.  8pwt.  to  Ib.  Ana.  a'o  lb. 

4.").     How   many  oz.  Troy  in  the    Brazilian    Emperor's 
diamond,  which  weighs  1680  carats?  Ans.  11  £  oz. 

46.  Reduce  *75  pwt.  to  3.  Ans.  30  3. 

47.  4gr.  to  5.  Ans.  glo  5- 

48.  19cwt.  to  oz.  Ans.  30400  oz. 

49.  221  dr.  to  T.  Ana.  .000044140625  T. 

50.  184-5  to  dr.  av.  Ans.  41}  dr. 

51.  96  oz.  av.  to  oz.  Troy.  Ans.  87  2  oz.  Tr. 

52.  How  many  wino  gal.  in  a  tank  3ft.  long  by  2j  ft. 
wide  and  lift,  deep?  Ans.  75??  w.  gal 

5'{,    How  many  U.  S.  bushels  in  a  bin  9.3ft.  long  by 
8|  ft.  wide  and  2!  ft.  deep?  Ans.  61  bu.,  nearly. 

M.     Reduce  21  bbl.  of  beer  (36  gal.)  to  bid.  of  wine, 
Coll  gal.)  Ana.  '20 -i 

55.     1  bu.  to  wine  gal.  Ans.  9.31  gal.,  nearly. 


150  RAY'S   HIGHER   ARITHMETIC. 


56.  1  bu.  to  beer  gal.  Ans.  7.G25 
Su QUESTION. — Reduce  to  cu.  in.,  and  then  to  gallons. 

57.  45  f  3  to  0.  Ans.  TVs  0. 

58.  2a  f  3  to  m.  Ans.  1200  m. 

59.  If  f  of  a  piece  of  gold  is  pure,  how  many  carata 
fine  is  it?  Ans.  207 

60.  In  181  carat  gold  what  part  is  pure,  and  what  par! 
alloy?  Ans.  j'l  and  3*3 

CASE    II. 

ART.  220.  To  reduce  a  compound  number  to  a  simple 
number  of  any  denomination. 

RULE. —  Commence  with  the  hiyJtest  denomination,  if  it  bf  Re- 
duction Descending  ;  wil/t  Ike  lowest,  if  it  be  Reduction  A.icendlnrj. 
Ileduce  those  of  that  denomination  to  the  denomination  required, 
and  during  this  reduction  add  in  at  the  proper  times  those  of 
the  other  denominations. 

NOTES. — 1.  If  the  denomination  required  lie  between  the  highest 
and  lowest  of  those  given,  reduce  part  of  the  compound  number  by 
Reduction  Descending,  and  the  rest  by  Reduction  Ascending,  and 
add  the  two  results. 

2.  The  numbers  added  in  must  be  of  the  same  denomination  aa 
those  to  which  they  are  added;  mistakes  can  be  avoided  by  marking 
the  denomination  of  each  number  as  it  is  obtained. 

Reduce  5  Ib.  2  oz.  13  pwt.  10  gr.  to  grains. 

OPERATION. 

SOLUTION.  —  Since    this     is  lb-    oz-    Pvvt-      Pr- 

Reduction      Descending,     com-  5      ^      13      10 

mence  at  the  51b.,  and  reduce  )b    OI 

it  to  GOoz. ;  the  '2  oz.  added  in  6  2  oz  =  5    2 

made  G'Joz.;   tliis  is  reduced  to  OQ 

pwt.,  nn<l  1:5  pwt,  added  in.  rnak-  "                    '£•     "'      ^ 

ing    1 '253  pwt.;    this    finally   is  l-OOpwt.=  5      - 

brought  to  gr.,  and    the   10  gr.     -  "* 

added,  producing  30082  gr.  for  5022 

the  answer.  2500 

lb.     01.     pwt.        CT. 

30082gr.  =  5  2  13  10 


._ 220.  What  if  Cnsc  2d  ?  Tho  rule?  If  the  required  de- 
noijjination  lies  between  the  highest  and  lowest  given,  what  is  necessary  ? 
VVb.it  care  must  bo  exercised  in  the  additions ? 


REDUCTION    OF   COMPOUiND   NUMBERS.  157 

Jlcducc  2  c\vt.  3  qr.  18  Ib.  to  tuna. 
SOLUTION.— This  being  Re-  OPERATION. 

ID. 


duction    Ascending,   commence     ey  z 
Kiih    18  Ib.    and    reduce   it  to 
.72  qr.,    making    3.72  qr.    alto-          . 
geilier;    this  is  reduced  to  .93 


jf  a  cwt.,  making  2.93  cwt.  in 
ll.;  and  this  finally  is  reduced 
to  .l40oT.  —  the  result  required- 
The  successive  quotients  might 


20 


18.00 


qr. 


3.72  =3qr.  181b, 


twt.  cwt.     qr         Ib 

2.930  =  2    3    18 


T.  cwt.    qr.       Ib. 

.1465=2    3    18 


be  put  in  the  form  of  common 
fractions,  if  it  were  desirable. 

For  convenience,  the  different  simple  numbers  that  compose  the 
compound  number,  are  set  in  a  column,  the  lowest  at  the  top,  nnd 
the  others  under  it  in  order,  so  that  each  quotient,  as  it  is  obtained, 
cnu  be  placed  beside  those  of  its  own  denomination.  If  any  of  the 
denominations  are  vacant,  a  cipher  must  be  placed  in  the  column 
to  correspond. 

Placing  each  quotient  beside  the  number  of  its  own  denomination 
Is  the  adding  in  required  by  the  rule;  just  as  the  2oz.  13pwt.  and 
10  gr.,  are  added  in  in  the  first  example. 

REMARK.  —  The  rule  maybe  stated  thus:  Reduce  each  of  the 
timple  numbers  that  compose  the  compound  number,  to  the  required  de- 
nomination, and  add  the  results — but  in  practice  it  is  not  so  convenient 
as  the  one  given. 

EXAMPLES  FOR  PRACTICE. 

1.  4  cu.  yd.  5  cu.  ft.  256  cu.  in.  to  cu.  ft. 
SOLUTION.  —  4cu.  yd.  5cu.  ft.  =  113cu.  ft.,  and  256 cu.  in.= 

25G  X  T72 8  cu.  ft.  =  .2"7  cu.  ft.  Ans.  113  A  cu.  ft. 

2.  1 7  mi.  3  fur.  38  rd.  to  rd.  Ans.  5598  rd. 

3.  8  rd.  1A  ft.  to  iu.  Ans.  1602  in. 

4.  43ft.  Sin.  to  :d.  Ans.  2J5g 

5.  9  mi.  22  rd.  10.6175ft.  to  fur.     Ans.  *72.566-f- 

toodKSTiON. — When  the  divisor  is  5 A,  30},  fcc.,  multiply  botfc 
dividend  and  divisor  by  the  denominator  of  the  fraction  (2,  4,  &c.), 
•ud  tltfi,  divide.  •  The  quotient  will  not  be  altered. 

6.  464yd.  2ft.  Si  in.  to  miles.         ^ns.  .26415+ 

RKVIKW.  —  220.  Explain  Example  1st.  What  convenient  form  it 
adopted  for  Reduction  AscemJing?  If  any  denomination  is  vacant,  what 
is  necessary  T  How  might  tho  rule  have  been  stated  more  simply? 


158  RAY'S   HIGHER   ARITHMETIC. 


7.  Irai.  3  fur.  7. 2  in.  to  yd.  Ans.  2420.2yd. 

8.  20  fathoms  4$  ft.  to  miles.  Ans.  .03378  mi. 

9.  17  yd.  3  qr.  2  na.  to  na.  AM.  286  na. 

10  5yd.  4  in.  to  ells  English.        Ans.  4.08  E.  Eng. 

1 1  44  E.  Flcm.  2  na.  2  in.  to  qr.        Ans.  132.72  qr. 
•'     12     1  qr.  2na.  1.785  in.  to  na.  Ans.  6.793  na. 

13.  29E.Flem.lqr.2lna.  to  yd.  Ans.  22.1  5625  yd. 

1 4.  75  yd.  3  qr.  2g  na.  to  feet.  Ans.  227.7  ft. 

15.  4  sq.  rd.  13  sq.  yd.  5  sq.  ft.  98  sq.  in.  to  sq.  in. 

Ans.  174482  sq.  in. 

16.  7sq.  ft.  120.54sq.  in.  to  sq.  yd.    Ans.  .8708  — 

17.  63  sq.  rd.  10$  sq.  in.  to  sq.  ft.       Ans.  l7loli|| 

18.  1650  A.  3  R.  24. 64  P.  to  sq.  miles. 

Ans.  2. 5795375  sq.  mi. 

19.  301A.  IR.  1843P.  to  sq.  ch.     Ans.  301 3.67 T3s 

20.  1  sq.  mi.  424  A.  to  perches.          Ans.  170240  P. 

21.  21  A.  35  P.  to  sq.  chains.    Ans.  212.1875  sq.  ch 

22.  leu.  yd.  24 cu. ft.  876cu.  in.  to  cu.yd.  Ans.  1U'*6 

23.  What  part  of  a  cord  of  wood  is  a  pile  71  ft.  long 
by  5ft.  wide  and  1|  ft.  high?  ADS.  foVf 

24.  83  cords  115  cu.  ft.  1600  cu.  in.  to  tuns  of  round 
timber.  Ans.  268g|gT. 

SUGGESTION. — First  to  cu.  ft.;  then  to  tuns  of  round  timber. 

25.  7  tuns   of  hewn   timber   32  cu.  ft.  480  cu.  in.   to 
cords.  Ans.  2-j3?i4  C. 

26.  31b.  7oz.  14.23!  pwt,  to  gr.     Ans.  20981. 6  gr. 

27.  19p\vt.  6gr.  to  Ib.  Ans  sYo  lb. 

28.  4  lb.  22]  gr.  to  oz.  Ans.  48Tg5o  oz. 

29.  77oz.  12pwt.  10.464gr.  to  !b.  Ans.  6.468483 

30.  68  lb.  81  oz.  to  gr.  Ans.  395808  gr. 

31.  49  carats  2T7egr.  to  oz.  Troy.        .4ns.  3sl  oz.  Tr. 

32.  A   diamond  weighing  3  oz.  16  pwt.    is   how   mat? 
carats  t  Ans.  57 •)  carats. 

33.  3ft  73  12.24 gr.  to  5.  Ans.  36.90053. 

34.  95  Ii9  to  gr.  Ans.  4350  gr. 

35.  23  23  14.1  gr.  to  3-  Ans.  .3G34'8  3 

36.  1  lb  1  H  gr.  to  3-  A  ns.  90  ,  «  o  .y 


REDUCTION   OF   COMPOUND  NUMBERS.  159 

37.    43  23  15.5gr.  to  ft.  Ans.  .051302-Mb. 

38  10  T.  9  cwt.  2  qr.  23  Ib.  to  Ib.       AM.  20973  Ib. 

39  3qr.  18  Ib.  15  oz.  10.  08  dr.  to  cwt. 

Ans.  .9397}$ 
40.    IT.  6  cwt.  Iqr.  24  Ib.  2oz.  4|dr.  to  Ib. 

Ans.  2649.  14^ 
4*1.     Iqr.  124  oz.  to  tuns.  AM.  .Ol28f|  T. 

42.  762  Ib.  8  oz.  3  dr.  to  cwt.        AM.  7.625T28  cwt. 

43.  121b.av.  9oz.lOdr.  toTroygr.  Ans.  88210]  sgr 

SDQ.  —  First  reduce  to  dr.;  then  to  gr.  by  27^.     (Art.  188.) 

44.  633313  15.232gr.  to  Ib.  av.  AM.  .442176. 

45.  5323  7gr.  to  pwt.  AM.  14£j  pwt. 


46.  15  Ib.  Troy,  11  oz.  4  pwt.  9.085gr.  todr.av. 

-4ns.  3356.  71168  dr.  av. 

47.  6  oz.  10.  48  pwt.  to  3.  4ns..  52.  1923. 

48.  37  gal.  2  qt.  1  pt.  3  gills  to  qt.          Ans.  ISOjqt. 

49.  4  gal.  2  pt.  to  gills.  Ans.  130  gills. 

50.  3  qt.  1  pt.  2.  3  gills  to  gal.  AM.  .9461  gal. 

51.  52  gal.  Iqt.  1.052  gills  to  pt.    Ans.  418.263  pt. 

52.  1  gal.  3  qt.  Is  gills  to  gal.  Ans.  li?  gal. 

53.  47  bu.  3  pk.  2  qt.  to  pt.  4ns.  3060  pt. 

54.  2  pk.  6  qt.  1  .  8  pt.  to  bu.  4ns.  .  71  ft  bu. 

55.  3bu.  2  pt.  to  pk.  Ans.  12  33  pk. 

56.  3  pk.  1.091  pt.  to  bu.  4ns.  .76  7  12  bu. 

57.  2  pk.  7  2  qt.  to  pt.  4/ts.  47  pt. 

58.  8bu.  3f  pk.  to  qt.  4ns.  286  qt. 

59.  286  bu.  3pk.  1  qt.  to  imp.  bu.        4ns.  278.02— 
Sua.—  First  to  U.  S.  bu.;  then  to  cu.  in.;  then  to  imp.  hu. 

60.  99  w.  gal.  1  qt.  1  pt.  to  imp.  gal.   4ns.  82.7904. 

61.  67  beer  gal.  3  qt.  1  .  94  pt.  to  dry  qt. 

4ns.  285.  32567— 

62.  56  w.  gal.  Iqt.  2g  gills  to  beer  qt.  4ns.  184.  603— 

63.  13bu.  Ipk.  7qt.l.35pt.tow.gal.  4ns.  125.  58— 

64.  27  gal.  3  qt.  1  pt.  of  beer  to  gills.     4ns.  108831 

BE  VIEW.—  220.   Is  it  as  convenient  in  thi*  form?    Who»  th«  divisor  U 
>  301,  Ac.,  what  should  he  done? 


50  RAY'S  IIIGIIER   ARITHMETIC. 


G3.  1 4  f  5  5  f  3  48  m  to  Cong.  Ans.  .115030  j'g  Cong. 

GG.  1  0  3f3  6K3  to  m.                       Aiia.  9510m. 

G7.  3f3  35m  to  fj.                                  Ans.  Ur%. 

G8.  20  7f5  4f3  58m  to  f5.          Ans.  39.62^3. 

GO.  2  yr.  108  da.  18  hr.  40  min.  to  sec. 

AM.  72470400  sec. 

70.  G  yr.  44  da.  8  hr.  35  min.  to  wk.      Ans.  319  vVr'a 

71.  21  hr.  4  min.  54. 6  sec.  to  da.  Ans.  .87840072da. 

72.  3  wk.  5  da.  12  hr.  26  min.  1U  sec.  to  hr. 

Ans.  636r7c°/olir. 

73.  74 da.  IGhr.  45  min.  23. 028 sec.  to  yr. 

Ans.  .204G525567,  nearly. 

74.  How  many  sec.  in  a  solar  year?  Ans.  3155G928. 

75.  9°  1 3. G"  to  minutes.  Ans.  540.226' 
7G.    42' 57V' to  degrees.                      Ans.  .715972° 

77.  163°  28'  7"  to  seconds.  Ans.  588487". 

78.  79  12°  6'  20"  to  degrees.  Ans.  222.10o° 

79.  4'  32.756"  to  circum.    Ans.  .00021046,  nearly. 

80.  $76  5ct.  21  m.  to  mills.  Ans.  760521 

81.  9  ct.  9.1054  mills  to  $.  Ans.  $.0991054 

82.  $391  7  mills  to  cents.  Ans.  39 100. 7  ct. 

83.  $84  82}  ct.  to  mills.  Ans.  84325'}  in. 

84.  20  eagles  $6  1  dime  2i  ct  to  $.          Ans.  $206| 
83.  4  eagles  3. 142  mills  to  dimes.     Ans.  400.03142 

86.  5  dimes  9  ct.  6  mills  to  eagles.  Ans.  .0596 

87.  £304  19s.  2Ad.  to  £.  Ans.  £304j§i 

88.  12s.  10]d.  tos.  Ans.  12.^s. 

89.  £58  7s.  lid.  to  d.  Ans.  14015  d. 

90.  £25  4|d.  to  s.  Ans.  500  !£s. 

91.  Reduce  1  cwt  to  pwt.  Ans.  29166J 

92.  How  many   ounces  of  gold  weigh  as  much   as 
pound  of  lead?  Ans.  14/2 

CASE   III. — TO  REDUCE  A  SIMPLE  NUMBER.  OP  ANY  DENOMI- 
NATION   TO   A   COMPOUND    NUMBER. 

ART.  221.    If  the  given  number  is  a  whole  number,  it 
may  be  reduced  to  a  compound  number  by  this 


REDUCTION    OF  COMPOUND  NUMBERS.  1(U 

RULE. — Reduce  the  given  number  to  the  next  higher  denomina- 
lion,  reserving  the  remainder;  reduce  the  quotient  to  (he  next 
ki<//ier  denomination,  and  reserve  the  remainder.  Continue  thus 
until  the  highest  denomination  has  been  reached,  or  until  the  quo- 
tient is  so  small  as  not  to  admit  of  further  reduction.  The  last 
quotient  with  the  several  remainders,  form  the  compound  number 
required. 

NOTE. — Each  remainder  is  of  the  same  denomination  as  tne 
dividend  from  which  it  is  obtained. 

REMARK. — The  operations  under  this  and  the  preceding  rule 
serve  to  prove  each  other. 

ART.  222.  If  the  given  simple  number  is  a  common 
or  decimal  fraction,  it  may  be  reduced  to  a  compound 
number  of  lower  denominations  by  this 

RULE. — Reduce  the  given  fraction  to  the  next  lover  denomina- 
tion, reserving  the  whole  number,  if  any,  of  that  denomination. 
If  there  is  a  fraction  in  this  result,  reduce  it  to  the  next  lower 
denomination,  reserving  the  whole  number,  if  any,  of  that  denomi- 
nation. Continue  this  process  until  no  fraction  occurs,  or  until 
the  lowest  denomination  has  been  reached.  The  whole  numbers 
reserved,  with  the  last  fraction,  if  any,  will  be  the  compound 
number  required. 

NOTE. — If  the  given  simple  number  be  a  mixed  number,  the 
whole  number  which  it  contains  may  be  reduced  to  a  compound 
number  by  the  first  of  the  rules  given  above,  and  the  fraction  by 
the  last  rule,  and  the  two  results  united. 

Reduce  1706  inches  to  a  compound  number. 

SOLUTION. — The  operation  is     12")l706  in 

similar  to  that  in  the  2d  exam-  — —  _ 

pic    Arl.  218,  except  that  each  *2  ft.  2  in. 

remainder  is  written  as  a  whole  47  yd.  1  ft.  2  in. 

Dumber  of  its  own  denomination, 

instead  of  a  fraction  of  the  next  higher  denomination;  as  142ft. 
1  in.  instead  of  142-f2  ft.,  and  47  yd.  1  ft.  instead  of  47g  yd. 

RRVIEW. — 221.  What  !s  Case  Sd?  What  is  the  rule  for  reducing  a 
simple  whole  number  to  a  compound  number?  How  is  the  denomination 
»f  eich  remainder  known?  How  are  operations  under  this  rule  verified? 
222.  What  is  the  rule  for  reducing  a  simple  fractional  number  to  a  com- 
pound number?  How  is  a  simple  mixed  number  reduced  to  a  compound 
number  ? 

14 


U)-->  RAY'S   HIGHER   ARITHMETIC. 


Reduce  5  of  a  wine  gallon  to  a  compound  number. 

SOLUTION.— Multiply  f  gal.  by  4;  the  w-|^  qj-    P*-     gill 

result  is  2$  qt. ;  cut  off   whole  number  2  —  -      1      1J 

as  part  of  the  compound  number  required.      ?.  0 

Multiply    5  qt.  by  2;  the  result  is  1;'  pt.:  qt.  2  I 

tut  off  the  1  pt.  and  reduce  the  3  pt.  to  1^  2 

fills.     This  being  the  limit  of  the   table,  pt.  1  | 

Hie  operation  must  stop,  and  the  compound  4 

number  required  is  2  qt.  1  pt.  1^  gills.  -11     -TT 

Reduce  4905.06185  Ib.  av.  to  a  compound  number. 


25)49^5  Ib. 

4)l_9_6_qr.&lb.  ^989 6"  o*. 

2 0)49  cwt.  1J3 

~~2~T.  9  cwt.  5  Ib.  1  5 . 8  3  3  6  dr.  Am. 

SOLUTION. — The  fraction  .OG185,  reduced  to  lower  denominations, 
is  15.8336  dr. ;  the  whole  number  4905  Ib.  reduced  to  higher  denomi. 
nations,  is  2  T.  9  cwt.  5  Ib.  These  results  united  give  2  T.  9  cwt.  5  Ib 
15.833(5  dr,  for  the  answer. 

Reduce  1657  yd.  to  a  compound  number. 

SOLUTION.  —  Divide   the  5  3")  165  7  vd 

1657  yd.  by  5^,  to  reduce  it  to  2 
rd.     To  do  this  conveniently, 

multiply  both  dividend  and  .  1  J  O  Q  1  *  nail  yd. 

divisor  by  2,  making  the  di-  40)301  rd.  3  half  yd. 

visor   11,    and    the    dividend  u  ,,        rt1      ,    -i       , 

8314  half-yards,  without  al-  J  fur'  21  rd"  l\  ^d' 

teiing  the  quotient     The  re-  But  ^  yd.  —  1  ft.  6  in. 

»mir,,lcr  3  is  also  half-yards  Am    ^  fur>  21  rd.  1  yd.  1  ft 

(Art.  221.  Note),  and  is  there-  f>  • 
fore  written  ^  yd.  =  1  .J  yd.; 

Ihe  i  yd.  is  then  reduced  to  a  compound  number  by  the  last  rule,  and 
j)»oed  to  the  result  already  obtained. 

REDUCE   TO    COMPOUND  NUMBERS, 

EXAMPLES.  ANSWERS. 

1.  44753Aft.     .     .     =  8  mi.  3  fur.  32  rd.  5  ft.  6  in. 

2.  99.75yd =18rd.  2ft.  3  in. 

3.  tfur =  33rd.  lyd.  2ft.  6  in. 


REDUCTION   OF   COMPOUND    NUMBERS.  103 

EXAMPLES.  ANSWERS. 

4.  8760531  in.      .    =  138  mi.  2  fur.  5  rd.  1  ft.  9  in 

5.  904.178fath =  904  fath.  1.068ft 

6     8871  na =  55  yd.  Iqr.  3]  na. 

7.  69.1525yd =  69  yd.  2.44  na. 

8.  GOliaqr =  150  yd.  1  qr.  I  na. 

9.  52. 003  E.  Fl =  52  E.  Fl.  .036  na 

10.  181 1.0625  sq.ft.     =201sq.  yd.  2sq.  ft.  9sq.in. 

11.  300027sq.  in.       =  231sq.  yd.  4sq.  ft.  75sq.  in. 

12.  64. 10826  P. =  1R.  24.10826  P. 

13.  832,'J  sq.  yd.      =  832  sq.  yd.  2  sq.  ft.  15?  sq.  in 

14.  1.10475  sq.  mi.     .     .      =  1  sq.  mi.  67  A.  6.4  P. 

15.  322. 7372  A.     .     .     .    =  322  A.  211.  37.952  P. 

16.  706.2814sq.  ch.  .     .    =70  A.  2  Pv.  20. 5024  P. 

17.  1567804  sq.  in.  =1209sq.yd.  6  sq.ft.  76sq.in. 

18.  87g  sq.ft.      .     .     .    =9sq.yd.  6  sq.ft.  90  sq.in. 

19.  93111  A.       .     .     .    =1  sq.  mi.  291  A.  211.  5  P. 

20.  535sq.  mi =  22  A.  2  R.  14  A  P. 

21.  |  sq.  rd =16  sq.  yd.  7sq.ft.  36  sq.  in. 

22.  T7T  cu.  yd =17cu.  ft.  314i*T  cu.  in. 

23.  1013854cu.in.  =21cu.yd.  19cu.ft.  1246cu.in. 

24.  .0038yr.     .     .      =  1  da.  9  hr.  l7uiin.  16.8  sec. 

25.  65.387cu.  ft.  =  2cu.  yd.  lieu.  ft.  668.736cu.  in. 

26.  4.2045cu.yd. 

=  4cu.  yd.  5cu.  ft.  901.1 52  cu.  in. 

27.  18.9142mi. 

=  18mi.  7 fur.  12rd.  2yd.  2ft.  11.712 in. 

28.  40152381  min.  (365|  da.  to  a  yr.) 

An».  7yr.  231  da.  14  hr.  38  min.  52Asec. 

29.  131IC =  13C.  120Scu  ft 

30.  8. 56^  T.  hewn  timber.     .     .      =  8T.  28$cu.  ft, 

31.  4000  gr.  Troy =  8  oz.  6  pwt.  16  gr 

32.  78?  Ib.  Troy.     .     =  78  Ib.  8  oz.  11  pwt.  10^  gr. 

33  *267.3pwt.     .     .    .     =llb.  loz.  7pwt.  7.2gr 

34  ISoz.  Troy =  15  pwt.  9^  gr. 

35  45.54oz.Tr.      .     =  31b.  9  oz.  10  pwt.  19.2gr. 

36  692  pwt =21b.  10  07.  12  pwt 


HJ4  KAY'S   HIGHER   ARITHMETIC. 

EXAMPLES.  ANSWERS. 

37.  13. 7G441b.   .    .   =  13  Ib.  9315  19  2.944  gr. 

38.  805H3 =  8ib.  43  53  19  9Jgr. 

39.  fas =55  19  lojgr. 

40.  905G29 =  314ft.  53  33  19- 

41.  ytb 8g  13  19  7T3Tgr. 

42.  lY0053.62gr.  Apoth.  =  29tb.  63  23  13.62gr. 

43.  467^9 =llb  Y3  33  29  12  gr. 

44.  .37013 =  192.206gr. 

45.  T85T.     .     .    =  10cwt.  2qr.  16lb.  lOoz.  lOjdr. 

46.  .4815  Ib.  av =  7oz.  11.264  dr. 

47.  809[Acwt.=40T.9c\vt.  3qr.l61b.  lOoz.  lOjdr. 

48.  4220311b =  211T.  1  qr.  61b. 

49.  7331  qr.    .      .     .  =9T.  3cwt.  Iqr.  18  Ib.  12  oz. 

50.  ilcwt =  2qr.  201b.  13oz.  5;!,dr 

51.  8 A  gal =8  gal.  3 Ji  gills. 

52.  10072  gills =314gal.  3qt. 

53.  95ijr|t.        .....        =  23  gal.   3qt.l?  gills. 

54.  301.46pt.    .     .   =37  gal.  2  qt.  1  pt.  1.84  gills. 

55.  808g  qt.  dry  measure     .     .      =  25  bu.  1  pk.  f  pt. 

56.  2191009. 3dr.=  4T.5cwt.2qr.81b.lOoz.  1.3 dr. 

57.  560052l43oz.av.=  ll7T.10cwt.31b.4oz.l3ridr. 

58.  365. 242241 4  days. 

Ans.  365  da.  5  br.  48  min.  49. 65696  sec. 

59.  £jg =6s.  78Td. 

60.  $33| =$33.625. 

61.  |\vk =4  da.  16  br. 

62.  .7  1'3 =5  f3  36m. 

63.  T9»5  br =  33  min.  45  sec. 

64.  ^  bu =1  pk.  n  pt. 

65.  .555£ =lls.  lid. 

CO.  f'igal =2qt.  Hpt. 

67.  67.7»)8 =£3.  7s.  0.12.1. 

(58.  27.35° =27°  21'. 

Hi).  32.4  O =4Cong.  6f3  3f3  12m 

70.  gCong «=8()  8fS  IfS  36m 


REDUCTION   OF    COMPOUND   NUMBERS.  165 

EXAMPLES.  ANSWERS. 

71     .075qt =.6  gill. 

72.  3.07pk =  3pk.  1.12pt. 

73.  46-18  f 3 «5f30f357m. 

74.  260234" =72°  17'  14". 

75.  1246.05' =20°  46'  3". 

76.  18563d .    .    =  £7  14s.  83d. 

77.  281923ss =  £140  19s.  lid. 

78.  .8054bu =  3  pk.  1  qt.  1.5456  pt. 

79.  100000m =1302f340m. 

80.  .1934  sign '.     .     =5°  48'  7. 2". 

81.  7543dimes =  $7.57ct.  5m. 

82.  I7.052pk =4bu.  Ipk.  .832pt. 

83.  2[?circum.       .     .     .      =  2c.  327b  16'  21TV'. 

84.  19019.2m.      .    .    .  =  20.  7f3  4f3  59.2m. 

85.  84312  mills. =$84.312. 

86.  cs  of  a  dollar =  44  ct.  6 /^  in. 

87.  TVof  an  eagle =$4.730fi 

88.  2093. 57  cents. =$20.9357. 

89.  795  of  a  degree =  28'  25  ft". 

90.  6407]  pt.  dry  meas.  .     .     =  100  bu.  3  qt.  l\  pt. 

91.  10808107. 87  sec. 

=  125da  2hr.  lomin.  7. 87  sec. 

92.  6. 045964  yr. 

=  6yr.  16  da.  18 hr.  38  min.  40.704  sec. 

93.  17000.12  da.  (365J  da.  to  a  yr.) 

AM.  46  yr.  198  da.  14  hr.  52  min.  48  sec. 

94.  22303. 5d.  Am.  £92  18s.  7Ad. 

REMARK. — If  our  tables  of  Weights  and  Measures  were  on  tin 
decimal  basis,  like  U.  S.  Money  and  the  French  tables,  the  same 
rules  and  methods  would  do  for  compound  as  for  simple  numbers. 
Common  fractions  would  also  occur  less  frequently,  the  compara- 
tively complicated  processes  that  arise  from  their  use  would  be 
avoided,  and  the  computations  required  in  ordinary  business  trans- 
actions would  be  much  shortened  and  simplified. 

As  it  is,  Compound  numbers  must  be  treated  somewhat  differ- 
ently from  Simple  numbers;  though  the  rules  and  operations  are 
not  entirely  new,  but  rather  modifications  of  those  already  ex- 
plained. 


100  RAY'S   HIGHER   ARITHMETIC. 


ADDITION  OF  COMPOUND  NUMBERS. 

AKT  223.  RULE. —  Write  the  numbers  to  be  added,  units  of 
the  same  denomination  in  a  column,  reducing  any  fraction*  to 
lower  denominations,  until  none  are  found  in  any  but  the  right- 
hand  ct'luitin.  Add  the  right-hand  column,  and  reduce  the  result, 
if  In  rye  enough,  to  the  next  higher  denomination;  write  the  re* 
mainder,  if  any,  under  the  column  added,  and  carry  tne  quotient 
to  the  next  column.  Add  the  next  column,  reduce,  xet  down,  and 
carry  as  before,  and  continue  so  until  all  the  column*  have  been 
added. 

PROOF. — Same  as  in  addition  of  simple  numbers. 

NOTE.  —  If  the  right-hand  column  contain  common  or  decimal 
fractions,  add  them  according  to  the  usual  rules ;  if  any  of  the 
higher  denominations  in  the  answer  has  a  fraction,  rtdace  it  to 
lower  denominations,  and  add  it  in. 

Add  3  bu.  2]  pk. ;  1  pk.  lA  pt. ;  5  qt.  1  pt. ;  2  bu.  1]  qt  ; 
and  .125  pt. 

SOLUTION. — Reduce    the     bu.   pk.  qt.      pt. 

fraction  in  each  number  to       322       0=3  bu.  2]  pk. 
lower   denominations,    (Art.  1      0       lj  =  1  pk.  lA  pt. 

222,)  and  place  units  of  the  51=  5  qt.    1     pt. 

same  kind  in  columns.     The       201          |  =  2  bu.  1£  gt. 
right-hand     column,     when  |  —  .125  pt. 

added,  gives  8,7?  pt,  =  1  qt        6      Q      1      13V=  Am. 
12  j  pt. ;  write  the  IVj  and 

add  the  1  qt.  with  the  next  column,  making  9  qt.  =  1  pk.  1  qt. ;  w-ite 
the  1  qt.  and  carry  the  1  pk.  to  the  next  column,  making  4  pk  = 
1  bu.;  as  there  are  no  pk.  left,  set  down  a  cipher  and  carry  1  bu. 
to  the  next  column,  making  6  bu. 

Add  2rd.  9ft.  7!  in.;  13  ft.  5. 78  in.;  4  rd.  lift.  6  ir  , 
Ird.  103ft.;  6rd.  14ft.  6|  in. 

SOLUTION. — The  numbers  are  pre-  '  ^   ~_ 

pared,  written,  and  added,  as  in  the  ^  ~  t'rra 

last  example;    the  answer  is   16  rd.  ^  ft 

9A  ft,  O.Goo  in.    The  i  foot  is  then  re-  ^        ^  * 

duced  to  <5  inches,  (Note),  and  added  *        |^ 

to  the  fl.C55  in.,  making  15.C55  in.  =  b-b- 


1  ft.  3.055  in.      Write  the  3.055  in.,      16  9\  9.655 

and  carry  the  1  ft.,  which  gives  1C  rd.  but     4  ft.  =  6. 

10  ft.  3.055  in.  for  the  final  answer.         -.  g       "-Tft 


ADDITION   OF   COMPOUND   NUMBERS.  18*7 

1.  Add  7  mi.;  3  fur.  261  rd.;  10  mi.  14  rd.  7ft.  Gin.; 
5.  24  fur.;  37  rd.  16ft.  2  Jin.;  1  mi.  12  ft.  8.720  in. 

AM.  12  mi.  4  fur.  20rd.  9ft.  4.  033!  in. 

2.  6.19yd.;    2yd.  2  ft.  9|  in.;    1  ft.  4.54  in.;  10  yd. 
2.376ft.;  I  yd.;  1|  ft.;  gin.   Ans.  21  yd.  2  ft.  3.517  in. 

3.  3yd.  2qr.   3  na.   liin.;    Iqr.  2|na.;    6yd.   1  aa 
2.175  in.;  1.63yd.;  5  qr.  ;  3  na. 

Ans.  12yd.  Ina.  0.755  in. 

4.  4  E.  Fr.  4  qr.  2  na.  ;  7  E.  Fr.  5  qr.  1  A  na.;  3  qr.  3  na. 
1  in.;  |E.Fr.;  l.bna.  ;  ^  in.      Ans.  14E.Fr.  3  na.  |  in. 

5.  2E.  Fl.  Iqr.  Una.;   5  E.  Fl.  3»  na.;  2qr.  23  na.; 
1E.F1.  2qr.  3.8na.;  gE.Fl.   Ans.ll  E.F1.  Iqr.  1  ft'ona. 

6.  losq.yd.    5  sq.  ft.    87  sq.  in.;    lOjsq.  yd.  ;    10  sq. 
yd.   7.  22  sq.ft.;    4  sq.  ft.   121.  6  sq.  in.;  Msq.  yd. 

Ans.  43  sq.  yd.  7  sq.  ft.  37r78  sq.  in. 

7.  101  A.  2  K.  18.  35  P.;    66  A.  1  R.  34A  P.;  20  A.; 
12  A.  2.84R.;  5  A.  13.33JP. 

-A«*.  205  A.  3R.  19.  781  P. 

8.  23  cu.  yd.  14  cu.  ft.  121  6  cu.  in.;    41  cu.  yd.  6  cu.  ft. 
642.132cu.  in.;    9  cu.yd.  25.065cu.  ft.;  7^  cu.yd. 

Ans.  75  cu.  yd.  4  cu.  ft.  1279.252  cu.  in. 

9.  eC.;  |cu.  ft.;  1000  cu.  in; 

Ans.  107  cu.  ft.  1072  cu.  in. 

10.  21b.Tr.6fd.;  Iftb.;    12.68pwt.;  lloz.  13pwt. 

gib.,  i£oz.;  6  pwt. 

Ans.  5  Ib.  Tr.  9  oz.  9  pwt.  2.  85  3  gr. 

11.  85l4.6gr.;   4.183;   7.523;   232318gr.;   13 
12  gr.;  |3.  AM.  lib.  2$  43  13. 
•12.     AT.;    9  cwt.  1  qr.  22  Ib.  ;    3.06qr.;    4  T.  8.764 
cwt.;  3qr.  6lb.;   T72  cwt.    Ans,  5  T.  6  cwt,  2  qr.  14:J7o  Ib. 

13.  .3lb.  av.;  Boz.;  ^jdr.  Ans.  5  oz.  GJ  T's  dr. 

14.  Cgal.  31  qt.;  2gal.  Iqt.  3.32gills;  1  gal.  2nt.  Apt.; 
|  gal.;   fiqt.;   I  pt.  yl«s.  11  gal.  2  qt.  .403  gill. 


15.  4  gal.  Skills;  10  gal.  3qt.  Upt.;  8  gal.  3  pt.  ;  5.6J 
gal.;  2.3qt.;  l'.27pt.;  fgill.  Aus.  29gal.  2qt.  .22jgill. 

RKVIEW.  —  222.  What  would  be  the  advantages  if  our  weights  and 
measure:)  wore  on  the  decimal  basis?  223.  Whiit  is  the  rulo  fur  addition  of 
compound  number."  V  Wlmt  is  the  proof?  If  the  ri^ht-hand  column  con- 
tains common  or  decimal  fractions,  what  must  be  done?  If  any  of  thl 
bibber  denominations  of  the  answer  has  a  fraction,  what  must  bo  donnl 


i(>S  RAY'S   HIGHER   ARITHMETIC. 

10.    Add  Ibu.  Apk.;  Tfli  bu.;  3Pk.  5qt.  1  !  pt.:  !>!..,  ;}.28 
pk.;  7<jt.   l.lGpt.;    ,-do  pk.       Ana.  12bu.  8pk.  .4G.'"|>t. 

17.  Add  |bu.;   fpk.;   gqt.;   apt.      AUK.  ii  pk.   ^  pt. 

18.  Add   6f3  2f3  25m;   2U3;    7f'3  ^^m;   1  f 3 
2lf3;  3f3  6  1-3  ol  m.  -4/m.  14  f 3  7  i'3  38m. 

ID.     AddAwk.;  ida.;   -I.hr.;   2  min.;  A  sec. 

-4«.s.  4  da.  30  min.  301  sec. 

20.  Add  3.2Gyr.  (365da.  each);  118  da.  5hr.  42  min. 
37  i  sec.;   63.4da.;  7s  hr.;    1  yr.  G2da.  lUhr.  24s  min.; 
T7^.,  da.  Aits.  4  yr.  340  da.  1  hr.  14  min.  55 -j  sec. 

21.  Add  27°  14' 55.24";    9°  18V;    1°  loj';    116° 
44' 23. 8"  Ann.  154°  14' 57.29" 

22.  Add  $84  let.   5.27m.;    67  ct.  8m.;    $25   Oct. 
23  in.;  79fiofadol.;  25^  ct.    .Am.  $110  60  ct.  5. 045  in. 

23.  $J;|ct.;gm.  Ans.  50  ct.  2g  m. 

24.  Add  $3  7m.;    $520ct.;    $100  2  ct.  Cm.;     $19 
let.  AHS.  $127  23  ct.  4.!  m. 

25.  Add  £21  Gs.  3 id.;  £517.?s.;  £9.085;  16s.  74!d.; 
£fsz.  ^l/i*.  £37  10s.  8.15d. 


SUBTRACTION  OF  COMPOUND  NUMBI^RS. 

ART.  224.  RULE. — Prepare  and  write  the  numbers  as  in  ad- 
lUion  of  compound  numbers,  placing  the  subtrahend  below.  Com- 
mence at  the  riyht,  and  proceed  to  the  left,  subtracting  each  lower 
number  from  the  one  above,  and  setting  the  remainder  below.  If 
a  lower  number  is  larger  than  the  one  above  it,  add  to  the  upper 
as  many  units  of  its  denomination  as  make  one  of  the  next 
higher;  subtract  and  carry  1  to  the  next  ji (jure. 

PROOF. — Same  as  in  subtraction  of  simple  numbers. 

NOTE. — If  fractions  .are  in  the  riglit-liand  column,  subtract  them 
by  the  usual  rules;  if  a  fraction  is  in  any  of  the  higher  denominations 
of  the  answer,  reduce  it  to  lower  denominations,  and  add  it  in. 

Subtract  1  yd.  2.45  ft.  from  9yd.  1  ft.  Gl  in. 

SOLUTION. — Change  the  A  in.  to  a  decimal,  yd.     ft.       ia. 

making  the  minuend  9  yd.  1ft.  6.5  in.;  reduce  9       1       G.5 

the  A't  ft.  to  inches,  making   the  subtrahend  1       2       5.4 

1  yd.  2  ft.  5.4  in.     To  subtract  2  ft.  from  the  T= p T~ T 

number  above,  add  3  ft.  ( =  1  yd.)  to  the  1  ft., 

making  4ft.;  set  the  remainder,  2ft.,  below,  and  to  compensate  for 


(SUBTRACTION   OF    COMPOUND   NUMBERS.  169 

the  3  ft.  added  above,  add  1  yd.  to  the  next  lower  figure,  vrhich  givei 
2yd.;  the  remainder  is  then  7yd.  and  the  answer  "yd.  2ft.  1.1  in. 

1.  Subtract  16  rd.  8  ft.  l.{  in.  from  23  rd.  1,2  ft. 

Ans.  6  rd.  9  it.  7.15  in. 

2.  fnii.  from  3  fur.  24.8Grd.  AHS.  16.86rd. 
3      1  35yd.  from  4yd.  2qr.  1  na.  1$  in. 

Ans.  3  yd    1  qr.  §  in. 

1.     2%  E.  Fl.  from  2-g  E.  E.        Ans.  1  yd.  1  qi    1  in. 
5      2  sq.rd.  24  sq.yd.  91  sq.in.  from  5  sq.rd.  16  sq.yd. 
ti'i  sq   ft.          Ans.  2  sq.  rd.  22  sq.  yd.  8  sq.  ft.  41  sq.  in. 

6.  .56i  sq.  yd.  from  7  sq.  ft.  18.27  sq.  in. 

Ans.  2  sq.ft.  3. 87  sq.in. 

7.  2R.  19iP.  from  11  A.    Ans.  10  A.  111.  20|P. 

8.  384  A.  IK.  3. 92  P.  from  1.305sq.  mi. 

Ans.  450  A.  3K.  28. 08  P. 

0.     'j?]cu.  ft.  from  i|cis.  yd.  Ans.  4cu.ft.  1503  cu.in. 
1C.     IS  cu.  yu.  25  cu.  ft.  1204.9  cu.  in.  from  20  cu.  yd. 
4cu.fl    lOOOcu.in.  Ans.  6cu.  yd.  5cu.  ft.  1523. leu.  in. 

11.  9.362oz.  Troy  from  lib.  15pwt.  4gr. 

Ans.  3oz.  7pwt.  22.24gr. 

12.  5  oz.  1 5  pwt.  from  3  Ib.  22  gr. 

Ans.  2  Ib.  6  oz.  5  pwt.  22  gr. 

13.  L§3  from  -frg.  Ans.  35  13  15^  gr. 

14  23196gr.from4gigr.  Ans.  33  55  13  14igr. 

15  56  T.  9cwt.  Iqr.  23 Ib.  from  75. 004  T. 

Ans.  1ST.  lOcwt.  2qr.  10 Ib. 

16  4fc\vt.  from  3  qr.  11  Ib.  14  oz.  103  dr. 

Ans.  2qr.  5  Ib.  15  oz.  6^  dr. 

17.  ^  Ib.  Troy  from  g'tlb.  uv.  Ans.  0. 

18.  3  gal.  2qt.  If  pt.  from  8  gal.  1.1  qt. 

Ans.  4  gal.  2  qt.  |  pt. 

19.  12  gal.  1  qt.  3  gills  from  31  gal.  li  pt. 

Ans.  18  gal.  3qt.  3  gills.    / 

20      .0625bu  from  3pk.  5  qt.  Ipt. 

Ans.  3  pk.  3  qt.  1  pt. 

REVIEW. — 224.  What  is  the  rule  for  subtraction  of  compound  numbers  T 
M  Lat  is  tho  proof?  If  fractions  are  in  the  right-hand  column,  what  must 
bo  Jone  »  V'hat,  if  a  fraction  is  in  any  of  the  higher  denominationi  of 
the  answer? 

15 


170  RAY'S  HIGHER   ARITHMETIC. 

21.  2  p.k.  .84  pt.  from  3bu.  41  qt. 

Am.  2bu.  2pk.  3qt.  l.GGpt. 

22.  15  wine  gal.  from  15  beer  gal. 

An*.  3}t  w.  gal.a=3w.  gal.  1  qt.  1 7?  gills. 

SUGGESTION. — Reduce  the  beer  gal.  to  vr.  gal.  (Art.  191.) 

23.  10  U.  S.  bu.  from  10  imperial  bu.  of  G.  Britain, 

AM.  Ipk.  2qt.  0.17+pt, 
SCUOESTION. — Reduce  the  imp.  bu.  to  U.  S.  bu.  (Art.  194). 

24.  If34f338m  from  4f32f3. 

Am.  2f5  5f3  22m. 

25.  .9  of  a  day  from  {wk.  AM.  201ir.  24min. 

26.  3  da.  16  br.  47  min.  33.3  sec.  from  1  wk. 

AM.  3  da.  7br.  12  min.  2G.7BCC. 

27.  275  da.   9  br.   12  min.   59  sec.    from   2.4816  yr 
(allowing  3G5j  days  to  tbe  year.) 

.4ns.  lyr.  265  da.  18  br.  29  min.  21. 16  sec. 

28.  1832  yr.  8  mon.  18  da.   from  1840  yr.  5  mon.  2C 
da.,  considering  1  mon.  to  be  30  days,  as  is  customary  in 
business  involving  time.  Ans.  7  yr.  9  mon.  8  da. 

29.  What  is  tbe  difference  of  time  between  Aug.  5th, 
1848,  and  Mar.  14th,  1851?       Ans.  2  yr.  7  mon.  9  da. 

SUGGESTION. — Proceed  as  in  last  example,  writing  Aug.  6th  in 
the  subtrahend,  as  7  mon.  5  da.,  and  March  14th  in  the  minuend, 
as  2  mon.  14  da.,  since  these  dates  are  respectively  that  long  aftei 
the  beginning  of  the  year;  allow  80  days  for  a  month. 

30.  Find   tbe   difference   of   time   between  Sept.  22d, 
1855,  and  July  1st,  1856.  -4ns.  9  mon.  9  da. 

31.  Between  Des.   31st,  1814,  and  April  1st,  1822. 

Ans.  7  yr.  3  mon. 

32.  Between  May  20th,  1855,  and  Oct.  15th,  1857. 

,4ns.  2  yr.  4  mon.  25  da. 

33.  Subtract  43°  18'  57.18"  from  a  quadrant. 

Ans.  46 °  41'  2.82" 

34.  17°  29'  from  24°  52".  Ans.  6°  31'  52". 

35.  161°  34'  11.8"  from  180°.  An*.  18°  25'  48.2" 

36.  $12.857  from  $19.  Ans.  $6.143 

37.  4 mills  from  $40.  Ans.  $39.996 


R  •  v  i  •  w. — 224.  How  is  the  difference  of  time  between  two  dates  found ) 


MULTIPLICATION    ^p  COMPOUND  NUMBERS         1 

38.  1  ct.  from  $1  and  1  m.  Ant.  99  ct.  1  m. 

39.  86  ct.  and  .6  mill  from  $2.62i  Ant.  $1.7644 

40.  .08  dime  from  let.. 8  mills.  Ant.  let. 

41.  fo  ct.  from  $32-  Ant.  8  ct.  4$  m. 

42.  $5  43  ct.  2j  m.  from  $12  6  ct.  85  m. 

Ans.  $6  63  ct.  5§  m. 

43.  £9  18s.  6R  from  £20.          Ans.  £10  Is.  5R 

44.  |s.  from  }25£.  Ans.  3gd. 


MULTIPLICATION  OF  COMPOUND  NUMBERS. 

ART.  225.    Since  every  compound   number  can  be  re 
duccd  to  a  simple  number  of  citber  of  its  denomination* 
(Art.  220),  tbc  multiplication  of  a  compound  number  will 
only  differ  from  tbo  multiplication  of  a  simple  number  by 
the  reduction  before  and  after  multiplying. 

GENERAL  RULE. 

TO   MULTIPLY   A   COMPOUND   NUMBER   BY   ANY   SIMPLE 
NUMBER   WHOLE   OR   FRACTIONAL, 

Reduce  the  compound  to  a  simple  number  of  either  of  its  de- 
nominations, and  multiply  as  in  simple  numbers.  The  product 
will  be  a  simple  number  of  the  same  denomination  as  the  mul- 
tiplicand, and  may  be  reduced  to  a  compound  number. 

NOTE. — It  is  generally  best  to  reduce  the  multiplicand  to  its 
lowest  denomination. 

Multiply  2  bu.  3pk.  7qt.  Upt.  by  10.8724. 
Bu.  pk.  qt    pt.  1  0.8  7  2  4 

2371*  =191|  pt. 

J.  27181 

llpk.  108724 

8  978516 

95  qt  108724 

2  2)2079.3465  pt. 

T 9l]  pt.  8)1 039  gt.  1.3465  pt. 

4)129  pk.  7qt. 
3  2  bu.  1  pk. 
An*.  32  bu.  1  pk.  7qt.  1.3465  pt. 


172  RAY'S  HIGHER  ARITHMETIC. 


ART.  226  If  the  multiplier  is  a  whole  number,  tho 
reductions  may  take  place  during  the  multiplication,  instead 
of  before  and  after  it. 

TO  MULTIPLY  A  COMPOUND  NUMBER  BY  A  SIMPLE  WHOLE 
NUMBER, 

ROI.E. — Legin  at  the  lowest  denomination;  multiply  each  of 
if,e  simple  numbers  that  compose  the  compound  number  in  suc- 
cession: reduce  each  product  to  the  next  higher  denomination, 
letting  the  remainder  below  the  number  multiplied,  and  carrying 
the  quotient  to  the  next  PRODUCT. 

PROOF. — Same  as  in  multiplication  of  simple  numbers 

NOTES. — 1.  If  the  multiplier  is  a  composite  number,  we  may 
multiply  by  its  factors  in  succession,  as  in  Art.  53. 

2.  AVhen    the  multiplications  and  reductions  can  not  be  readily 
performed  in  the  mind,  do  the  work  on  one  side,  and  transfer  the 
results. 

3.  If  there  be  a  fraction  in  the  lowest  denomination  of  the  multi- 
plicand, multiply  it  first;  if  one  occurs  in  any  of  the  higher  denomi- 
nations of  the  product,  reduce  it  to  lower  denominations,  and  add 
it  in. 

Multiply  9hr.  14  min.  8.17  sec.  by  10. 

SOLUTION. — Ten  times  8.17sec.  hr.    min.      sec. 

=  81.7 sec.  =  1  min.  21.7 sec.     Write  9      14      8.1*7 

21.7  sec.  and   carry    1  min.  to   the  ^  10 

140  min.  obtained  by  the  next  mul-  q     ~OA      91 91 h~~ 

tiplication.     This  gives  141  min.  = 

2hr.  21  min.     Write  21  min.  and  carry  2hr.     This  gives  92  hr.  = 
3  da.  20  hr. 

Multiply  12  A.  3R.  28|  P.  by  84. 

SOLUTION. — Since  84  =  7  X 12,  multiply  '        '         ' 

by  one  of  these   factors,  and   this  product  -*--1  "      ~j~s 

by  the  other;    the  last  product  is  the  one  '__ 

required.     The  same  result  can  be  obtained  90  1      38g 

by  multiplying  by  84  at  once;   performing  12 

lio  work  on  one  side  and  transferring  the  -.  r\or  q~ 
results. 

REVIEW. — 225.  What  is  the  rule  for  multiplying  a  compound  number 
by  a  simple  number,  whole  or  fractional  ?  To  which  of  its  denomination! 
should  tho  multiplicand  generally  be  reduced?  226.  What  is  tho  rule  for 
multiplying  a  compound  number  by  a  simple  whole  number  ?  The  proof? 


MULTIPLICATION    OF   COMPOUND   NUMBERS. 


1.     Multiply  7rd.  10ft.  5  in.  by  6.  Ans.  45  rd.  13ft. 
2      2 mi.  3 fur.  27 rd.  by  8.     Am.  19mi.  5fur.  IGrd 

3.  16  yd.  2|  in.  by  21.        Ans.  33*7  yd.  1  ft.  3]  in. 

4.  1  uii.  14  rd.  8|  ft.  by  97. 

^HS.  101  mi.  3 fur.  6rd.  8ft.  Sin. 

5.  4yd.  2ft.  9. 14  in.  by  47A. 

Ans.  231yd.  2ft.  9.73T6Tin. 

6.  12E.F1.  3ina.  by  18.  Ans.  221E.F1. 

7.  6  E.E.  4qr.  3.44na.  by  28. 

Ans.  195  E.E.  Iqr.  .32na 

8.  5  sq.  yd.  8  sq.  ft.  106  sq.in.  by  13. 

Ans.  77  sq.  yd.  5  sq.ft.  82  sq.in. 

9.  41  A.  3R.  26. 1087  P.  by  9.046. 

Ans.  379  A.  23. 4593+ P. 

10.  10  cu  yd.  3  cu.ft.  428.15  cu.in.  by  67. 

Ans.  678  cu.yd.  1  cu.ft.  1038.05  cu.in. 

11.  7  oz.  16  pwt.  5|  gr.  by  174. 

^n«.  113  Ib.  3oz.  5  pwt.  16Agr. 

12.  231913  gr.  by  20.  Ans.  6  g  3  3. 

13.  16cwt.  Iqr.  7.88lb.  by  11. 

Ans.  ST.  19cwt.  2qr.  11.68lb. 

14.  7  Ib.  6oz.  12?  dr.  by  283.44. 

Ans.  1  T.  1  cwt.  3  Ib.  15  oz.  8.98?  dr. 

15.  1  qt.  3j  gills  by  7.  Ans.  2  gal.  2  qt.  3  gill. 

16.  5 gal.  3  qt.  1  pt.  2  gills  by  35.108. 

Ans.  208  gal.  1  qt.  1  pt.  2.52  gills. 

17.  26  bu.  2pk.  7qt.  .37pt.  by  10. 

Ans.  267  bu.  7qt.  1.7pt. 

18.  3f348mbyl2.  Ans.  5f  g  5f 3  36  m. 

19.  18  da.  9hr.  42min.  29. 3  sec.  by  16T7i. 

Ans.  306  da.  4hr.  25  min.  2  sec.,  nearly. 

20.  $1072  9ct.  2m.  by  424.  Ans.  $454567  8m. 

21.  £215  16s.  2jd.  by  75.  Ans.  £16185  14s.  |d. 

22.  10°  28'  42£"  by  2.754.  Ans.  28°  51'  27.765" 

REVIEW. — 226.  If  the  multiplier  is  a  composite  number,  what  may  bt 
done  ?  When  tho  multiplying  and  reducing  can  not  readily  be  performed 
In  the  mind,  what  should  be  done  ? 


174  RAY'S   HIGHER   ARITHMETIC. 


ART.  227.  The  difference  of  time  between  two  places 
is  4  hr.  18min.  20  sec. :  what  is  their  difference  of  lon- 
gitude? 

SOLUTION. — Every  hour  of  time  cor-          hr.      min.        sec. 
icsponds    to    15°  of    longitude;    every  4         18         20 

minute    of    time    to    15'  of   longitude;  1 5 

every  second  of  time  to  15"  of  longi-        rTT^      o  rj/        oTy/ 
tude,    (Art.  201).      Hence,   multiplying  ^  ^  Lonff 

the  hours  in  the  difference  of  time  by 

15  -will  give  the  degrees  in  the  difference  of  longitude,  multiply- 
ing the  minutes  of  time  by  15  will  give  minutes  (x)  of  longitude,  and 
multiplying  the  seconds  of  time  by  15  will  give  seconds  (")  of  lon- 
gitude; since  reducing  seconds  to  min.  and  min.  to  hr.  are  the  same 
as  reducing  (")  to  (')  and  (')  to  (°),  the  divisor  iu  both  cases  being 
always  CO,  hence, 

TO   CHANGE   DIFF.  OF   TIME    INTO   DIFF.  OF   LONGITUDE. 

RULE. — Multiply  the  difference  of  time  by  15,  according  to  the. 
rule  for  Compound  Multiplication,  and  mark  the  product  °  '  " 
instead  of  hr.  min.  and  sec. 

REMARK. — The  work  can  be  shortened  by  cancellation,  for  15 

X*  y  op. 
times  26  divided  by  CO  =ir_±— -=  6^'  =  6'  30",  since  £'  =  30". 

90  4 
Write  the  30"  and  cairy  the  6'.    Then  15  times  18  divided  by  60 

a=ffiX18  =  4i°==4°  30',  since  ^°  =  30/;  add  in  the  6'  to  be  car- 

00    4 

ried  with  this  30',  making  36'.  Carry  the  4°  to  the  next  product, 
60°,  making  64°;  the  answer  is  64°  36'  30",  as  before. 

1.  When  it  is  4  o'clock  P.  M.  at  New  York,  it   is  3 
hr.  18  rnin.  28. 4  sec.  r.  M.  at  Cincinnati:  the   longitude 
of  New  York  is  74°  V  0"  W. :  what  is  the  longitude  of 
Cincinnati?  Ans.  84?  24'  W. 

SUGGESTION. — Of  two  places,  the  one  having  later  time  is  cait 
of  the  other;  the  one  having  earlier  time  is  west  of  the  other. 

2.  When  it  is  1  p.  M.  at  St.  Louis,  it  is  8  hr.  14miD 
55  j  sec.   P.  M.  at  the  Cape  of  Good   Hope:  the  longitudi 
of  the  latter  place  is  18°  28'  45"  E.:  what  is  the  longi- 
tudo  of  St.  Louis?  Ans.  90°  15'  10"  W. 


R  E  v  i  E  w.  —  227.  What  is  tho  rule  for  converting  difference  of  time  into 
difference  of  longitude?  Illustrate  and  prove  it.  Show  how  tho  opera- 
tions under  thin  rule  can  be  shortened  by  cancellation. 


DIVISION   OF   COMPOUND  NUMBERS.  175 

3.  A  man  travels  from  Halifax  to  Chicago;  his  watch 
shows  9  A.  M.,  while  tho  time  at  Chicago  is  7  hr.  24  min. 
24  J  sec.  A.  M.     Tho  longitude  of  Halifax  being  63°   36' 
40"  W.:  what  must  be  the  longitude  of  Chicago? 

Am.  87°  30'  30"  W. 

4.  When  it  is  10  A.  M.  at  Stockholm,  it  is  3  hr.  24 
miu.  58  sec.  A.  M.  at  Wheeling:  the  longitude  of  Whec-1- 
ing  is  80°  42'  W.:    what  is  the  longitude  of  Stockholm? 

Ans.  18°  3'  30"  E. 

5.  Noon  comes  47  min.  17  sec.  sooner  at  Detroit  than 
at  Calvcston,  whose  longitude  is  94°  47'  15"  W.:  what 
is  the  longitude  of  Detroit?  Ans.  82°  58'  W. 

G.  Time  is  7  hr.  57  min.  2G|scc.  later  at  St.  Peters- 
burgh  than  at  New  Orleans,  and  the  longitude  of  the 
former  is  30°  19'  46"  E.:  what  is  the  longitude  of  the 
latter?  Ans.  89°  V  50"  W. 

7.  When   it  is  1    r.    M.   at  Utica,  whose   longitude  is 
75°  13'  W.,    it   is  11  hr.   52  min.  4  sec.  A.  M.  at  Little 
Hock:  what  is  the  longitude  of  the  latter? 

Ans.  92°  12'  W 

8.  When  it  is  3  P.  M.  at  Regent's  Park,  London,  it  is 
9  hr.  46  min.  31.2  sec.  A.  M.  at  the  University  of  Virginia, 
whose  longitude  is  78°  31'  29"  W.:  what  is  the  longitude 
of  the  former?  Ans.  9'  17"  W. 

9.  When   it  is  midnight  at   Madras,  in  India,  it  is  1 
hr.  23  min.  16. 2  sec.  r.  M.  at  Buffalo;  tho  longitude   of 
tho  former  place  is  80°  15'  57"  E.:  what  is  the  longitude 
of  tho  latter?  Ans.  78°  55'  W. 

10.  When   it  is  1  A.  M.  at  Constantinople,  it  is  11  hr. 
loinin.  25  j7s  sec.  P.  M.  of  the  previous  day  at  Paris,  and 
the  longitudo  of  Paris  is  2°  20'  22"  E.:  what  is  that  of 
Constantinople?  Ans.  28°  59'  E. 

11.  A  ship's  chronometer,  set  at  Greenwich,  points  to 
4  hr.   43  min.  12  sec.  P.  M.j   the  sun    on   the  meridian 
what  is  the  ship's  longitude?  Ans.  70°  48'  W. 


DIVISION  OF  COMPOUND  NUMBEHS. 

ART.  228.   Division  of  compound  numbers  like  division 
of  bimplo  numbers,  has  two  cases: 


17(5  RAY'S   HIGHER   ARITHMETIC. 


CASE  I. 

ART.  229.  To  divide  a  compound  number  by  a  simple 
number. 

Since  every  compound  number  can  be  reduced  to  a 
simple  number  of  cither  of  its  denominations,  (Art.  220), 
the  division  of  a  compound  number  by  a  simple  number 
will  only  differ  from  the  division  of  simple  numbers  by 
the  reduction  before  and  after  dividing. 

GENERAL    RULE 

FOR   DIVIDING   A    COMPOUND   NUMBER   BY   ANY   SIMPLE 
NUMBER,  WHOLE  OR    FRACTIONAL. 

Reduce  the  compound  number  to  a  simple  number  of  either  of 
its  denominations;  divide  as  in  simple  numbers:  the  quotient 
will  be  a  simple  number  of  the  same  denomination  as  the  divi- 
dend, and  may  be  reduced  to  a  compound  number. 

NOTE. — It  is  generally  best  to  reduce  the  dividend  to  its  lowest 
denomination. 

Divide  17  da.  5  hr.  24  min.  19.208  sec.  by  8. Of 
SOLUTION.— 17  da.   5hr.  24  min.   19.208  sec.  =  1488259.208  sec. 
which  divided  by  8.07  gives  a  quotient   1 84418.737  -|-  sec.,  and  thir 
reduces  to  2dn.  3hr.  13  min.  38.737-}- sec.,  the  quotient  required. 

ART.  230.  If  the  divisor  is  a  whole  number,  the  re 
ductions  may  take  place  during  the  division,  instead  of 
before  and  after  it. 

TO    DIVIDE   A    COMPOUND    NUMBER   BY    A   SIMPLE   WHOLE 
NUMBER. 

RULE. — Divide  that  part  of  (he  dividend  which  is  oflhe  highest 
denomination  first,  and  set  the  quotient  below:  reduce  ilie  remain- 
der, if  there  is  one,  to  the  next  lower  denomination,  add  in  those 
of  that  denomination  in  the  dividend,  and  divide  again.  Con- 
tinue so  until  the  lowest  denomination  has  been  used;  when,  if 
there  is  a  remainder,  it  should  he  expressed  as  a  common  or 
decimal  fraction  of  that  denomination. 

R  F.  v  i  K  w. — 228.  How  many  cases  in  division  of  compound  numbers? 
229.  What  is  tho  1st  case?  What  is  the  general  rule  for  dividing  a  com- 
pound number  by  a  simple  number,  wholo  or  fractional?  To  which  of  its 
denomination*  is  the  dividend  generally  reduced?  230.  If  the  divisor  i; 
a  simple  whote  nuuioer,  what  is  the  rule? 


DIVISION   OF   COMPOUND  NUMBERS.  177 


PROOF. — Same  as  in  division  of  simple  numbers. 
NOTE. — If  the  divisor  is  a  composite  number,  we  may  divide  by 
its  factors  in  succession,  as  in  Art.  66. 

REMARK. — This  rule  may  be  used  when  the  divisor  has  a  com 
mon  or  decimal  fraction,  by  multiplying  both  numbers  by  (he  de- 
nominator of  this  fraction,  which  will  convert  the  divisor  into  a 
whole  number,  and  yet  will  not  alter  the  quotient,  (Art.  75). 

Divide  106  Ib.  9oz.  14|  dr.  of  sugar  equally  amosg  8 
men. 

SOLUTION. — 8  into  106  Ib.  gives  Ib.       oz.      dr. 

a  quotient  13  Ib.,  and  21b.  =  32oz.  8)106      9      14| 

to  be   carried  to  the  9  oz.  making        .  _  -,  q      e          o~4 

•  ,.  -/IMS.      —    J.  O        O  05 

41  oz.;  8  into  41  oz.  gives  a  quotient 

6  oz.,  and  1  oz.  =  16  dr.  to  be  carried  to  the  14§  dr.,  making  30§  dr  • 

8  into  30§  dr.  gives  3§  dr.  and  the  operation  is  complete. 

If  $42  purchase  67  bu.  2  pk.  5qt.  If  pt.  of  meal,  how 
much  will  $1  purchase? 

SoLUTiON.-Since  42  =  6X7,   di-  *£  T%>  &  P<- 

vide  first  by  one  of  these  factors,  and  0;  O  <  ^  0  14 

the  resulting  quotient  by  the  other ;  the  7)11  1  0  iff 

laat  quotient  will  be  the  one  required.  :; ^  o  TL2  3~ 

A  man  travels  1472  mi.  6 fur.  32 rd.  in  59  days;  how 
much  a  day  docs  ho  average? 

mi.      fur.     rd.    mi. 

59)1472  6  32(24 
118 

292 
236 

SOLUTION. — As  the  divisor  is  5  6  mi. 

larger  than  12,  and  can  not  be  8 

separated   into    suitable   factors,  A^AP       />7  P 

proceed   as  in  1st  example,  per-  d.1  Q. 
forming  the  work  by  long,  in- 

•lead  of  short  division.  4 1  fur. 

40 

1672rd.(28lird. 
118 

492" 
472 

Am.  24  mi.  7  fur.  28|j}  rd.         20 


RAY'S   HIGHER   ARITHMETIC. 


1.  Divide  1C  mi.  2  fur.  29  rd.  by  7. 

Am.  2  ini.  2  fur.  27  rd. 

2.  37  rd.  14ft.  11. 28  in.  by  18. 

Ans.  3rd.  1ft.  8. 9 Gin. 

3.  43  E.  1-1.  Iqr.  S^na.  by  33.  ^H«.  1  E.F1.  3H  na. 

4.  675  C.  114. GO  cu.  ft.  by  83. 

^LHS.  80.  18.3453  +  cu.ft. 

5.  10  sq.  rd.  29  sq.  yd.  5  sq.  ft.  94  sq.  in.  by  17. 

Ans.  19  sq.  yd.  4  sq.  ft.  ll$H«q.in. 

6.  1000A.  by  160.  Ans.  GA.  III.. 

7.  Gsq.  mi.  35 P.  by  22i     Am.  170  A.  2H.  285  P. 

8.  1245  cu.  yd.  24  cu.  ft.  lG27cu.  in.  by  11.303 

Am.  110  cu.  yd.  G  cu.  ft.  338.4-f-cu.  in. 

9.  88  Ib.  IGpwt.  I7.62r.  by  54. 

Am.  lib.  7oz.  llpwt.  10.1+gr. 

10.  35  75  18gr.  by  12.  'Am.  23  19  16A  gr. 

11.  COOT.  7cwt.  86 Ib.  by  29. OG 

Am.  20  T.  13cwt.  20  Ib.  14  oz.  124- dr. 

12.  62  Ib.  av.  8  oz.  by  9G.  Am.  10  oz.  G  J  dr. 

13.  312  gal.  2qt.  Ipt.  3. 36  gills  by  72^ 

Am.  4  gal.  Iqt.  1.794- gills. 

14.  19302  bu.  by  6.215 

Am.  3105  bu.  2pk.  6  qt.  1.54-pt. 

15.  53bu.  Apt.  by  63.      Am.  3  pk.  2qt.  1.85— pt. 

16.  76  yr.  108  da.  2  hr.  38  min.  2G.18scc.  by  45. 

Am.  lyr.  254 da.  27niin.  31.25— sec. 

17.  19  hr.  53?  sec.  by  71 

Am.  2hr.  26  min.  16. 06-}- sec. 

18.  152°  46'  2"  by  9.  Am.  16°  58'  265" 

AHT.  231.  Since  the  difference  of  time  between  two 
places,  multiplied  by  15,  gives  their  difference  of  longitudo, 
the  product  being  marked  °  '  "  instead  of  hr.  min.  and 
eoc.:  conversely, 

TO   CHANGE   DIFF.  OP   LONGITUDE    TO   DIFP.  OP   TIME, 

RIT.E. — Divide  the  difference  of  longitude  by  15,  according  tc 
the  rule  for  Compound  Division,  and  mark  the  quotient,  hours, 
minutes,  and  seconds,  instead  of  °  '  ' 


DIVISION   OP  COMPOUND   NUMBERS.  170 


NOTE. — The  division  required  by  the  rule  can  be  shortened  by 
canceling,  in  a  manner  similar  to  that  explained  in  Art.  227. 

The  difference  of  longitude  between  two  places  is  81° 
89'  22";  what  is  their  difference  of  time? 

15)81°       _39^_      22" 

5hr.     26min.  31^  sec. 

SOLUTION. — 15  into  81°  gives  5  (marked  hr.),  and  6°  to  be  car- 
ried. Instead  of  multiplying  6  by  CO,  adding  the  39'  and  then 
dividing,  proceed  thus:  15  into  G°  is  the  same  as  15  into  G  X  GO/  = 

G  X  60-1 
— J-=24/,  and  as  15  into  39'  gives  2'  for  a  quotient  and  9' 

d  p 
remainder,  the  whole  quotient  is  2G'  (marked  min.),  and  remainder 

9  X604 

9' =  9  X  GO",  which  divided  by  15  gives  -^^-  =  36",  which  with 

-ip 

1  y75  obtained  by  dividing  22"  by  15  gives  37T73",  (marked  sec.) 
The  ordinary  mode  of  dividing  will  give  the  same  result,  and  may 
be  used  if  preferred. 

1.  What  time  at  Columbus  (long.  83°  3'  W.),  when  it 
is  4  P.  M.  at  Baltimore,  (long.  7G°  37'  W.)? 

Ans.  3  hr.  34  min.  16  sec.  P.  M. 

2.  What  time  at  Copenhagen  (long.  12°  34'  57"  E.), 
when  it  is  10  P.  n.  at  Mobile,  flong.  88°  11'  W.)? 

Ans.  4  hr.  43  min.  3s  sec.  A.  M.  the  day  after. 

3.  What  time  at  Pittsburg  (long.  79°  58'  W.),  when  it 
is  3  A.  M.  at  Dublin,  (long.  6°  20'  30"  W.)? 

Ans.  10  hr.  5  min.  30  sec.  p.  M.  the  day  before. 

4.  When  it  is  noon  at  Louisville  (long.  85°  30'  W.), 
what  time  at  Bangor,  (long.  G8°  47'  W.)? 

Ans.  1  hr.  6  min.  52  sec.  P.  M. 

5.  When  it  is  6  P.  M.  at  Havana  (long.  82°  22'  21" 
W.),  what  time  is  it  at  Paramatta,  (long.  151°  1'  35"  E.)? 

Ans.  9hr.  33  min.  35  j  5  sec.  A.  M.  the  day  after. 

6.  What  time  at  Cambridge,  Eng.,  (long.  5'  21"  E.) 
when  it  is  9  P.  M.  at  Cambridge,  Ma'ss.,  (long.  71°  7'  21 
W.)?         Ans.  1  hr.  44  min.  50^  sec.  A.  M.  the  day  after. 

RETIEW. — 230.  What  is  tho  proof?  If  tho  divisor  is  a  composite 
number,  what  may  bo  done?  How  can  this  rulo  bo  used  when  tho  divisor 
ho-s  a  common  or  decimal  fraction?  X31.  What  is  tho  rulo  for  converting 
difference  of  longitude  into  difference  of  time?  Illustrate  and  prove  it. 
How  can  tho  operations  under  tho  rule  bo  shortened? 


180  RAY'S  HIGHER   AKITHMETIC. 

7.  When  it  is  7  A.  M.  at  Washington  (long.  77°  1'  30 
W.),  what  time  at  Mexico,  (long.  99°  5'  W.)? 

Ant.  5  hr.  31  min.  46  sec.  A.  M. 

CASE  II. 

ART.  232.  To  divide  one  compound  number  by  anothef 
similar  compound  number,  the  quotient  being  an  abstract 
cumber. 

RULE. — Reduce  both  dividend  and  divisor  to  simple  number* 
of  the  same  denomination,  and  then  divide. 

PROOF. — Same  as  in  division  of  simple  numbers. 

NOTE. — It  is  generally  best  to  reduce  the  numbers  to  their  lowest 
denomination;  if,  after  reduction,  one  or  both  contain  a  fraction, 
proceed  as  directed  in  division  of  decimal  or  common  fractions. 

How  often  can  a  keg  of  2  gal.  3  qt.  li  pt.  be  filled  from 
a  barrel  of  molasses  containing  8*7$  gal.? 

SOLUTION.— 37A  gal.  =  300 pt.  and  2  gal.  3qt.  1^  pt.  =  23 J  pt.; 

then  300pt. ^234 pt.  =  300-f-^-=300X  — =  --=125  times.  Ans. 

3  /0      7 

1.  How  many  lunar  months  of  29  da.  12  hr.  44  min. 
2. 84  sec.,  in  a  solar  year,  (Art.  196,  Note)? 

Ans.  12.368-f- 

2.  How  many  steps,  2  ft.  9  in.  each,  will  a  man  take 
in  going  3.'  miles?  Ans.  6240. 

3.  The  wheels  of  a  locomotive  are  10ft.  5  in.  in  cir- 
^/  ?umfcrence,  and  make  8  revolutions  a  second ;  how  soon 

'  will  it  run  100  miles?  Ans.  1  hr.  45  min.  36  sec. 

4.  The   new  half-dollar  of  the  U.  S.  contains  7  pwt. 
4s  gr.  pure  silver;    how  many  dollars  in  11  oz.  2  pwt.  of 
pure   silver,  and   if  it  be  coined  into    66s.  what   is   one 
shilling  worth  in  U.  S.  currency? 

Ans.  $15.411,  and  Is.  =  23777>V  :t. 

5.  How  many  half  eagles,  each  weighing  5  pwt.  9gr, 
»nd  21  car.  2|gr.  fine,  can  be  made  of  1000  sovereigns, 
each  weighing  5  pwt.  3,274  gr.,and  22  car.  fine? 

Ans.  973IJIS 

REVIEW. — 232.  What  ia  tho  rule  for  dividing  ono  compound  number 
by  another  similar  one  ?  The  proof?  To  which  denomination  is  it  best 
to  reduce  tho  numbers  ? 


ALIQUOT   PARTS.  1&] 


6.  A  comet  moves  8°  17'  22j"  in  ono  day;  in  -what 
tiuic  will  it  complete  the  circuit  of  the  heavens,  or  300°  ? 

Ans.  43  da.  10  hr.  1C  ruin.  19  sec.,  ncai'ly. 

7.  How  many  persons  can  receive  each  $2  18ct.  7 2m., 
out  of  a  fund  of  $59  6|  ct?  Ans.  27. 

8.  How  many  half-crowns,  each  worth  2s.  6d.,  are  in 
£18  7s.  1043d.?  Ans.  147T29ff 

9.  The  Julian  calendar  assumed  the  year  365  da.  6hr., 
instead  of  365  da.  5  hr.  48  min.  48  sec.,  its  true  length; 
in  how  many  years  was  a  day  gained?        Ans.  128?  yr. 

10.  In  how  many  years  is  a  day  gained  by  the  Grego- 
rian calendar,  which  allows  for  the  fraction  of  a  day  by 
adding  in  97  days  in  400  years?  Ans.  3600  yr. 


ALIQUOT  PARTS. 

ART.  233.  Aliquot  parts  is  a  useful  method  of  finding 
a  product,  when  one  or  both  of  the  factors  is  a  compound 
number.  The  following  is  an  example  of  the  sort  of 
problems  to  which  it  is  generally  applied. 

What  cost  28  A.  3R.  25  P.  of  hnuUt  $16  per  acre? 

SOLUTION. — Multiply  $16,  the  price  $ 

of   1A.,  by  28;    the  product  $448  is  16 

(lie  price  of  28  A.     3  R.  is   made  up  28 

of  2R.  and  1  R.;  the  former  is  3  of  1  9  Q 

an  A.,  and  the  latter  4  the  former;  o^ 
obtain  the  price  of  2  R.  by  taking  i 


$448 
8 
4 
2 


.50 


the  price  of  1  A.,  or  4  of  $16  =  $8 ; 

the  price  of  1  R.  is  £  of  this  =  $4.         2  R.  =  ^ 

25  P.  is  equal  to  20  P.  and  5  P.;  the          1  R.  =  I 

former  is  3  of  1  R.,  and  worth  2  of       2  0  P.  =  | 

$4  =  $2 :  the  latter  being  £  of  20  P.,          5  P.  =  | 

is  worth  -}  of  $2  =  $A  =  50 ct.  These  &46'?  50 

results  added,  give  the  value  of  28  A. 

8  11.  25  P.,  =  $462.50.     The  same  result  could  be  obtained  by  re 

•lucing  28  A.  3  R.  25  P.  to  acres,  vii:  28.90625  A.,  and  multiplying 

it  l>y  16. 

ART.  234.  This  method  can  be  applied  when  the  mul- 
tiplicand is  a  compound  number,  as  in  the  following  ex- 
ample : 


182  RAY'S   HIGHER   ARITHMETIC. 


A  travels   3  mi.  5  fur.  1C  rd.  in  1  hr.j  how  far  will  he 
go  in  G  da.  9  hr.  18  min.  48  sec.,  (12  hr.  to  a  day)? 

m'  fur"    rd 


ex.-.in- 

ple  is  solved  like  the  pre- 
ceding, except  that  the  mul-  ^_ 

tiplications    and    divisions  da.                hr.1      33      0     24 

•re    performed   on   a  com-  6  =  8x9;264      4      32 


pound  instead  of  a  simple  1  5  min.  •• —  % 

number.  3      ..      =  1 

One  of  the  most  valuable  3  0  sec.  =  g 

applications  of  aliquot  parts  ^  5             • —  ^ 

is  when  the  product  is  to  be  g            — .  4 
U.  S.  money ;  for  instance, 


7    14 
1    18! 
9* 


D 


If  $47.52  is  paid  for  the  use  of  money  1  yr.,  how  much 
ought  to  be  paid  for  using  it  4yr.  7  mon.  19  da.  ? 

SOLUTION.  —  When    money    is  $47.52 

paid  for  the  use  of  money,  1  month  4 

is  reckoned  30  days,  and  the  aliquot 
parts  taken  accordingly.  The  mills 
in  the  result  are  usually  neglected 


j.mon.= 


190.08 

23.76 
3.96 
2.376 
.132 


if  they  are  under  5 ;  but,  if  they  are     -i  o  j        _<; 

6  or  over,  they  are  counted  1  cent.        1  An      -J 

The  above  result  would  be  called 

$220.31.  $220.308 

REMARK. — When  the  number  of  which  parts  are  taken,  ends  in 
0,  the  simplest  way  is  to  take  the  tenths,  instead  of  halves,  fourths,  &c. 
In  the  example  above,  since  1  mon.  =  30  da.,  separate  19  da.  into 
18  da.  and  1  da.;  the  former  is  yfi(j  of  30  da.,  and  the  value  corres- 
ponding is  found  by  multiplying  the  value  for  1  mon.  (3.9G)  by 
y\5  =  .0,  which  is  the  same  as  to  multiply  by  6,  and  set  U^  figures 
of  the  product  2.376  one  place  farther  to  the  right. 

ART.  235.  In  all  questions  in  aliquot  parts,  one  of 
'be  numbers  indicates  a  rate,  and  the  other  is  a  compound 
number  whose  value  at  this  rate  is  to  be  found. 

RULE  FOR  ALIQUOT  PARTS. 

Multiply  the  number  indicating  the  rate  by  the  number  of  thai 
denomination  for  whose  unit  the  rate  is  given,  and  separate  the 


REVIKW.— 233.   What  is  Aliquot  parta  ?    Explain  its  use.     *«i    What 
kind  of  a  number  may  tho  multiplicand  be? 


ALIQUOT  PARTS.  183 


numbers  of  the  oilier  denominations  into  parts  whose  values  can 
be  obtained  directly  by  a  simple  division  or  multiplication  of  one 
of  the  preceding  values.  Add  these  different  values ;  the  result 
will  be  the  entire  value  required. 

NOTE. — Sometimes  one  of  the  values  may  be  obtained  by  adding 
or  subtracting  two  preceding  values  instead  of  by  multiplying  OT 
dividing. 

EXAMPLES  FOR  PRACTICE. 

1.  If  a  person  travel  4  mi.  5  fur.  10  rd.  12  ft.  4  in.  in 
Ihr.,  how  far  will  lie  travel  in  7  hr.  37min.  28  sec.? 

Ans.  35 mi.  4  fur.  6rd.  2ft.  22?  in. 

2.  What  cost  86yd.  3qr.  2  na.  of  cloth   at  $2.43! 
per  yard?   (Turn  |  into  a  decimal.)          -4ns.  $211.76 

3.  Find   the   cost   of  231  A.   IK    34 P.   of  land   at 
$17.28  per  A.  AM.  $3999.672 

4.  What  is  the  cost  of  127yd.  of  carpet  at  $1.87-J 
per  yd.?  Ans.  $238. 122 

SOLUTION. — Here    the    only  127 

compound  number  is  the  rate  ex-  1   8  7  ' 

pressed  in  Federal  money.    Take 
it  as  the  multiplier. 


$127 


The  cost  of  127yd.  at  $1  per      5  0  ct.  =  2         63.50 
yd.  is  $127 ;  by  taking  suitable      2 5  ct.  =2         31.75 
parts  of  this,  the  cost  of  127yd.      12^ct.=  2         15.87J; 
is  found  at  60  ct.,  25  ct.,  12^  ct.,  a  $238.122 

yd.  respectively.'  The  sum  of  all 

these  is  the  cost  of  127yd.  at  $1.872  a  yd.  Therefore,  aliquot  parts 
can  be  used  to  find  the  value  of  any  number  of  articles,  when  the 
value  of  one  is  known  in  U.  S.  money,  by  finding  their  cost  at  $1  a 
piece,  and  taking  such  aliquot  parts  of  this  as  are  necessary  to  make 
the  cost  at  the  given  price. 

5.  Find  the  cost  of  42  cu.  yd.  24  cu.  ft.  of  earth  at 
$1.25acu.yd.  Ans.  $53.61 

6.  Of  7lb.  8oz.  16pwt.  11  gr.  of  gold  at  $15.46  an 
oz.  Ans.  $1435.04 


REVIEW. — 234.  What  is  one  of  its  most  valuable  applications?  Give 
an  example.  235.  In  questions  in  aliquot  parts  what  relation  exists  be- 
tween the  two  quantities?  What  is  the  rule  for  aliquot  parts?  How  can 
a  value  sometimes  be  obtained?  How  can  the  value  of  any  number  of 
wticlos  be  found  when  th«  price  of  one  ii  given  in  U.  8.  money? 


184  RAY'S  HIGHER  ARITHMETIC. 


7.  Find  the  cost  of  6  T.  13cwt.  2  qr.  21  Ib.  of  su-sir, 
at  |4.68|  a  cwt.  Am.  £626.77 

8.  Of  3  Ib.  7  oz.  of  cheese,  at  15  ct.  a  Ib.  Am.  5 1  -,",.  ct. 

0.    lyd.  of  cloth  is  worth  3  qt.  1  pt.  2  gills  of  wino: 
what  is  47yd.  2  qr.  1  na.  worth?  Ans.  44gal.  2qt.  2|gill. 

10.  In  1  wine  gallon  are  231  cu.  in.:  how  many  en.  in 
in  24  gal.  3qt.  1  pt.  2i  gills?  Ans.  57tf4|j 

11.  In  1   beer  gallon  are  282  cu.  in.:  how  many  cu.  in 
in  38  gal.  1  qt.  1  pt.  of  beer?  Am.  10821? 

\'l.  In  1  bushel  are  2150. 42  cu.  in.:  how  many  cu.  in 
in  15  bu.  1  pk.  6  qt.  1  pt?  Ans.  33230. 7  4- 

13.  What  is  the  value  of  29  gal.  2  qt.  1  pt.  of  wine,    at 
$2.25  a  gal.?  Ans.  $66.65f 

14.  What  is  the  value  of  46  gal.  1  qt.  If  pt.  of  beer,  at 
80 ct.  a  gal.?  Ans.  $13.94  4- 

15.  What  is  the  value  of  10  bu.  3  pk.  5  qt.  of  corn,  at 
62ict.  a  bu.?  Ans.  $6.82  — 

16.  If  £3    6s.  silver  weigh   lib.  Tr.,  how  much  will 
weigh  17  Ib.  11  oz.  16  pwt.  9  gr.?          Ans.  £59  7s.  4- 

17.  Built   8rd.  14ft,  10  in.   of  fence   in    1   da.;    how 
much  can  I  build  in  3  wk.  5  da.  9  hr.  46min.,  if  Ida.  = 
10  hr.,  and  1  wk.  =  6  da.?  Ans.  213rd.  6ft.  1  in.,  nearly. 

18.  If  a  man  is  2  hr.  25min.  38  sec.  in  digging  a  cu. 
yd.  of  earth,  how  long  will   he  be  in  digging  44  cu.  yd. 
22  cu.  ft.?  Ans.  108  hr.  46  min.  3lM  sec. 

19.  A  comet  moves  24°  6'  49"  in  1  hr. ;  how  far  will 
it  go  in  6hr.  14  min.  52  sec.?    Ans.  150°  39'  23. 3" 4- 

20.  A  pendulum  beating  54000  times  a  day,  beats  how 
often  in  4  da.  3  hr.  20  min.  5  sec.?  Ann.  223503s  times. 

ART.  236.  Aliquot  parts  can  be  applied  to  making  out 
bills  in  U.  S.  money,  when  the  prices  of  the  items  are 
given  in  State  currencies. 

What  cost  33  £  gal.  of  wine,  at  14s.  7-;d.  a  gal.,  New 
England  currency? 

SOLUTION.— The  tables  of    State  $332  =  $33.50 
currencies,  (Art.  207),  show  that  in  2 


the    New    England   States,  6s.  =  §1.  ^o8    =  O 
Tho  cost  at  6s.  or  $1  a  gal.  is  $33.60, 

from  which,  by  multiply  ing  and  taking  f>  j 

suitable  parts,  the  cost  at  12s.,  2s.,  6cl.,  -i  i  j' 


$67 
11.167 
2.792 
.698 


1  ;,<!.  is  found  ;  the  sum  of  these  is  the  ''*~~Q~I — 7*~c 

•nut.  nt  14a    lli\      us  romiiroii.  §81.66 


oost  at  14s.  7^dM  as  required. 


ALIQUOT  PARTS.  185 


What  cost,  in  New  England  currency, 

1.  35yd.  cloth,  at  7s.  6d.  a  yd.?  Am.  $43.75 

2.  19|  yd.  of  muslin,  at  2s.  4d.  a  yd.?  Ans.  $7.68 

3.  26  caps,  at  8s.  6d.  a  piece?  Ans.  $36.83 

4.  lOJdoz.  copy  books,  at  4s.  3d.  a  doz.?    Ans.  $7.44 

5.  158 Ib.  starch,  at  9d.  a  lb.?  Ans.  $19.75 

6.  451b.lOoz.butter,at2s.  8d.  per  lb.?  Ans.  $20.28 

Sec  OESTION.— 45  lb.  10  oz.  =  45f  lb.  =  45.625  lb. 

7.  34  da.  work,  at  5s.  4d.  a  day.?  Ant.  $30.22 

8.  18 doz.  and  8 eggs,  at  Is.  lOd.  a  doz.?    Ant.  $5.70 

9.  72 bu.  3pk.  corn, at  3s.  3d.  a  bu.?    Ant.  $39.41 
10.  861  yd.  carpet, at  10s.  5d.  a  yd.?     Ant.  $150.82- 

ART.  237.    What  cost,  in  New  York  currency, 

1.  14|  yd.  of  calico,  at  Is.  2d.  a  yd.?      Ans.  $2.15 

2.  12  bbl.  potatoes,  containing  2A  bu.  each,  at  4s.  6d 
»bu.?  Ans.  $16.87^ 

3.  2  bu.  3  pk.  6  qt.  of  dried  peaches,  at  17s.  8d.  a  bu.? 

Ant.  $6.49 

4.  4 gal.  Iqt.  Ipt.  oil,  at  2s.  3d.  a  gal.?    Ans.  $1.23 

5.  33  dictionaries,  at  5s.  6d.  a-piece?     Ans.  $22.68.f 

6.  49  boxes  matches,  at  4jd.  a-piece?       Ans.  $2.30 

A.RT.  238.    What  cost,  in  Pennsylvania  currency, 

1.  3qt.  1  pt.  molasses,  at  2s.  a  qt.?          .4ns.     93  ct. 

2.  1  box  candles  (40  lb.),  at  Is.  8d.  a  lb.?  Ant.  $8.89 

3.  12  lb.  of  coffee,  at  lOd.  a  lb.?  Ant.  $1.33 

4.  9  gal.  2  qt.  Ipt.  of  milk,  at  6d.  a  qt.?  Ans.  $2.57 

5.  4  wk.  5  da.  wages,  at  13s.  4d.  a  week?  Ans.  $8. 38 

6.  23rd.l05ft.offencing,at20s.ard.?^7W.  $63.03 

ART,  239.    What  cost,  in  S.  Carolina  currency, 

1.  13  gal.  3qt.  of  oil,  at  3s.  6d.  a  gal.?  Ans.  $10.31i 

2.  A  ham  of  I7ilb.,  at  lid.  a  lb.?  Ans.  $3.43f 

3.  43  lb.  of  butter,  at  Is.  4ld.  a  lb.?  Ans.  $12.67 

4.  161  yd.  of  silk,  at  8s.  3d.  a  yd.?  Ans.  $29.83 

10 


186  RAY'S   HIGHER   ARITHMETIC. 


ART.  240.     What  cost,  in  Canada  currency, 

1.  7  gal.  1  qt.  of  honey,  at  6s.  lOd.  a  gal.?  Ant.  $9.91 

2.  5bu.  Ipk.  7qt.  of  dried  apples,  at  16s.  8d.  a  bu.? 

Ans.  $18.23 
ART.  241.      MISCELLANEOUS  EXAMPLES. 

1.  How  long  is  a  rope  winding  276  times  round  a  tre<J, 
whose  circum.  is  4yd.  2ft.  6tin.?     Ans.  1339yd.  4  in. 

2.  What  is  the  area  of  a  field,  length  67  rd.  8  ft.  5  in., 
breadth  39  rd.  11  ft.  2  in.?         Ans.  16  A.  2  R.  38+  1'. 

3.  How  many  bbl.  (3l£  w.  gal.)  in  a  room  22ft.  3  in. 
long,  16ft.  6  in.  wide,  lift.  4 in.  high?      Ans.  9884V 

i  4.  How  many  bu.  in  a  bin,  8ft.  10  in.  long,  4ft.  6  in. 
wide,  3  ft.  2  in.  high?  Ans.  101  bu.  5  qt.,  nearly. 

5.  How  much  land  in  a  rectangular  field,  88. 44  eh. 
long,  56.27ch.  wide?  Ans.  469  A.  2R.  2. 7008  P. 

G.  A  bought  a  pipe  of  wine  (137  gal.),  lost  9  gal.  2  qt. 
lipt.  by  leakage,  and  sold  the  rest  at  $2.372  per  gal.: 
how  much  did  he  receive?  Ans.  $302.37 — . 

7.  How  many  quart,  pint,  and  half-pint  bottles,  of  each 
an  equal  number,  can  be  filled  out  of  a  cask  containing 
44 gal.  2qt.  Ipt.?  Ans.  102. 

8.  A  man  can  mow  in  1  day,  2  A.  3  R.  20  P  of  grass: 
in   what   time   will    he    mow  78  A.  1  R.    36  P.,   allowing 
10  hr.  to  a  day?  Ans.  27  da.  2  hr.  57^J  min. 

9.  What  is  the  value  of  161b.  7  oz.  12  pwt.  3  gr.  of 
gold,  at  $15.85  an  oz.?  Ans.  $3163.76. 

10.  If  1  cu.  ft.  of  water  weigh   62j  lb.,  what   is   the 
"Weight  of  the  water  in  a  room  20  ft.  long,  15  ft.  5  in.  wide, 

9  ft.  10  in.  high?  Am.  94  T.  14  cwt.  3  qr.  21-11  lb. 

11.  If  a  ship  sail  10  mi.  6  fur.  18|rd.  per  hour,  how 
long  will  it  be  in  going  3236  mi.  2  fur.  36.508rd.? 

Ans.  12  da.  11  hr.  28  min.  6+ s*c. 

12.  In  a  rectangular  field,  are  160  A.  2  R.  36  P.;  one 
side  is  74.1 8  ch.:   what  is  the  other  ?      Ans.  21.67—  ch. 

13.  How  thin  is  a  cu.  in.  of  gold,  beaten  so  as  to  covci 
ft  space  46ft.  10  in.  by  41ft.  8  in.?        Ans.  SST'OTJO  in. 

14.  When    I   arrived   at  Cincinnati,  my  watch,  which 
kept  time  correctly,  was  42  min.  fast :  from  which  direc- 
tion  had  J   come,  and   how  far  in    that  direction    had    J 
traveled?  Ans.  From  the  east;  564J  mi 

NOTE. — A  degree  of  longitude  at  Cincinnati  is  about  53  £  mi 


RATIO.  187 


XII.  RATIO. 

ART.  242.  Ratio  is  a  Latin  word  signifying  relation  or 
connection ;  in  Arithmetic,  it  means  the  relation  of  one 
number  to  another,  expressed  by  their  quotient. 

The  ratio  of  2bu.  to  5bu.  is  f;  of  10yd.  to  3yd.  is  y30;  showing 
that  5  bu.  are  f  of  2  bu.,  and  3  yd.  T30  of  10  yd. 

Ratio  exists  only  between  quantities  of  the  same  kind, 
since  only  such  can  be  divided,  one  by  the  other. 

Since  a  ratio  is  a  quotient,  it  is  an  abstract  number, 
(Art.  60),  showing  how  many  times  one  number  contains 
the  whole  or  part  of  another. 

ART.  243.  The  ratio  of  two  numbers  is  indicated  by 
writing  them  in  the  order  in  which  they  are  mentioned, 
with  a  colon  (:)  between  them. 

The  ratio  of  4  to  9  is  written  4:9;  of  2]  to  4.65,  is  written 
2 1  :  4.65;  of  2  ft.  8  in.  to  1  yd.- 1  ft.,  is  written  2  ft.  8  in.  :  1  yd.  1  ft. 

Each  number  is  called  a  term  of  the  ratio,  and  both 
together  a  couplet  or  ratio.  The  first  term  of  a  ratio  is 
the  antecedent,  which  means  going  before;  the  2d  term  is 
the  consequent,  which  means  following. 

A  simple  ratio  is  a  single  ratio  of  two  terms;  as,  3  :  4=3. 
A  compound  ratio  is  the  product  of  two  or  more  simple 

ratios;  as,  4  X  5  :  3  X  7= is  the  product  of  the  sim- 
ple ratios  4:3  =  1  and  5:7=?. 

The  value  of  a  ratio  depends  not  on  the  absolute,  but 
on  the  relative  size  of  its  terms. 

TO   FIND   THE   VALUE   OP    ANY   RATIO, 

RULE. — Express  Hie  terms  in  the  same  denomination;  take  tht 
consequent  as  the  numerator,  and  the  antecedent  as  the  denomi- 
nator of  a  common  fraction ;  this  fraction  reduced  to  its  simplett 
form,  will  be  the  ratio  required. 

R E  v i E  w.— 2-12.  What  is  the  meaning  of  Ratio?  What  is  ratio  in 
Arithmetic?  Giro  examples.  When  can  two  quantities  have  ratio? 
What  kind  of  a  number  is  every  ratio?  Why?  243.  How  is  a  ratio  indi. 
eated?  Give  examples. 


188  RAY'S    HIGHER  ARITHMETIC. 


NOTE.  —  The  French  method  of  obtaining  the  ratio  has  been 
adopted  here.  The  English  method  makes  the  antecedent  the  nu- 
merator, and  the  consequent  the  denominator  of  the  fraction;  the 
ratio  of  Sin.  to  7  in.,  by  the  French  method,  is  f;  by  the  English,  |. 

Since  the  value  of  a  ratio  is  equal  to  the  consequent 
divided  by  the  antecedent,  it  follows  that 

The  antecedent  is  equal  to  the  consequent  divided  ly  th* 
value  of  the  ratio;  and  that 

The.  consequent  is  equal  to  the  antecedent  multiplied  by  the 
value  of  the  ratio. 

Hence,  if  the  value  of  a  ratio  is  known,  and  one  of  its 
terms,  the  other  can  be  found. 

What  is  the  ratio  of  9  to  15? 
SOLUTION  —  9  :  15  =  '-g5  =  f  =  if  Ans. 
What  is  the  ratio  of  2  A.  3  R.  25  P.  to  1  A.  ? 
SOIBTION.—2A.   3R.  25P.  :  1  A.  or  465  P.  :  160P.,  =  ^§§  = 
|f  Ans. 

What  is  the  ratio  of  4f  to  2i?  Ans.  $. 

oj 

SUGGESTION.  —  Here  the  rule  gives  a  complex  fraction  -f>  which 

4ff 
is  reduced  by  Art.  132. 

What  is  the  ratio  of  7.108  to  9.26f? 

0  %?         27  8 
SOL.-7.108  :  9.26|  =  = 


FIND   THE   RATIO 

1.  Of  7to5;  of  9tol;  of  2to4;  oflatoS;  of  36to50; 
of  112  to  16;  of2|to4i;  of  7itol3|;  of91to6£;  of  20 
tol|;  of  8-&to4$. 

Aiis.  *;  i;  2;  4;  ITS;   4;  lg;  Iff;  TO?;   iV;    iftW. 

2.  Of  2§  :  4|,  of  6.5  :  .013,  of  9f  :  17.28,  of  116J  : 
18.75,   of4i  :  9.8,  of  I  :  H,  of  i  :  i,  of  10.  08  :  31, 
of  2.176  :  14.3,  of  6.37A  :  34,  of  91  :  44.4. 

ll,  ,h,  11,  /r,  2A,  1258,  li,  27254,  6IH,  51,  4|. 


R  E  TI  E  w.—  213.  What  is  each  of  the  numbers  called  ?  What  arc  both 
together  called?  "Which  is  the  antecedent?  Which,  the  consequent  ?  Why 
10  called?  What  is  a  simple  ratio?  A  compound  ratio?  What  does  th« 
rain*  of  a  ratio  depend  on  ? 


RATIO.  189 

3.  Of  2  ft.  6  in.  to  3  yd.  1  ft.  10  in.  Ant.  4] 

4.  Of  4 mi.  6 fur.  20  rd.  to  Imi.  2 fur.  16 rd.  Am.  ]$| 

5.  Of  13  A.  3  R.  25  P.  :  6  A.  2  R.  10  P.      Am.  ft 

6.  Of  20.  12  cu.  ft.  :  150.  AM.  7U 

7.  Of  3  Ib.  10  oz.  6  pwt.  10^  gr.  :  2  Ib.  1443  pwt. 

AM.  TYB<£ 

»      8.    Of  2  g  3  3  1 3  :  5  g  7  3  14.32  gr.  Ans.  2^2& 
9.    Of  13  Ib.  :  9  Ib.  15.2  dr.  Ans.  |§8 

10.  Of  14  T.  12  cwt.  1  qr.  18.44  Ib.  :  7  cwt.  4i  Ib. 

Ans.  75 55 sis 

11.  Of  3  qt.  If  gills  :  8  gal.  1  pt.  4n«.  1012% 

12.  Of  10  gal.  1.54pt.  :  7 gal.  2qt.  .98pt.  Ans.  3841 

13.  Of  56  bu.  2  pk.  1  qt.  :  35  bu.  3  pk.  6.055  qt. 

Ans.  f|83 

14.  Of  5hr.  26min.  443  sec.  :  3|da.      Ans.  14|fT? 

15.  Of  2  yr.  22  da.  :  7  yr.  216  da.  Ans.  3f  fff  £ 

16.  Of  42°  15'  27^  :  90°.  Ans.  28708S5T 

17.  If  the  antecedent  is  7  and   the  ratio   lj,  what  is 
the  consequent?  Ans.  10 3 

18.  If  the  consequent  is  $13.42!  and  the  ratio  f ,  what 
is  the  antecedent?  Ans.  $35.80;i 

19.  If  the  antecedent  is  2|  and  the  ratio  6.048,  what 
is  the  consequent?  Ans.  15.7248 

20.  If  the  consequent  is  6  yd.  2  ft.  83  in.  and  the  ratio 
is  3s,  what  is  the  antecedent?  Ans.  2yd.  2i  in. 

21.  If  the  antecedent  is  5bu.  1.68pt.  and  the  ratio 
is  5g,  what  is  the  consequent?  Ans.  29 bu.  Ipk.  2qt.  j75£t. 

22.  If  the  antecedent  is  24.075  and  the  ratio  is  .1664, 
what  is  the  consequent?  Ans.  4.00608 

23.  If  the  consequent  is  4  and  the  ratio  f ,  what  is  the 
antecedent?  Ans.  lj| 

24.  If  the  consequent  is  27  Ib.  5oz.  14  dr.  and  the  ratio 
ii  |,  what  is  the  antecedent?      Ans.  45 Ib.  9oz.  12ifdr. 

25.  If  the  consequent  is  $7.43|  and  the  ratio  2a,  what 
ii  the  antecedent?  Ans.  $3.18.f 

REVIEW. — 243.  What  is  the  rule  for  finding  tho  value  of  a  ratio?  IIow 
many  methods  of  valuing  a  ratio?  Explain  the  difference.  What  is  the 
antecedent  equal  to?  Tho  consequent?  Hotr  is  tho  ratio  of  two  mixed 
numbers  found?  How  is  the  ratio  of  two  decimals  found? 


190  RAY'S   HIGHER   ARITHMETIC. 

2G.  Find  tie  value  of  $2.  56}  :  $10,  of  37ict.  :  33'  ct., 
of  $13.00;  :  $44,  of  $22  :  22  ct.  Ann.  343T)  S,  si,  TOO 

ART.  244.  Two  ratios  may  be  formed  with  the  same 
two  numbers,  by  taking  each  of  them  in  succession  as  the 
standard  of  comparison  ;  thus,  the  ratio  between  5  and  7 
is  i  or  7  ;  thn  former  is  the  ratio  of  5  to  7,  and  the  latter, 
the  ratio  of  7  to  5. 

One  of  the  ratios  which  can  be  formed  with  two  num^ 
bcrs  will  be  an  improper  fraction,  and  the  other  a  proper 
fraction  ;  for  the  sake  of  distinction,  call  the  former  the 
increasing  ratio,  and  the  latter  the  decreasing  ratio.  If  the 
two  quantities  are  equal,  the  ratio,  which  ever  way  it  is 
formed,  will  be  equal  to  1,  and  therefore  neither  increas- 
ing nor  decreasing,  but  a  ratio  of  equality. 

TO   MAKE   AN    INCREASING   OR   DECREASING   RATIO, 

RULE.  —  Write  the  tioo  numbers  in  the  form  of  an  improper 
fraction,  to  express  an  increasing  ratio;  but  in  the  form  of  a 
proper  fraction,  to  express  a  decreasing  ratio. 

1.  Make  an  increasing  ratio  with  7  and  18,     with  5i 
and  4f,     with  3  yd.  2  ft.  and  3  yd.  1ft.  5  in.,    with  6|  and 

6Qsi  An<,     18     22      132      1276 

.00.  A.ns.    ?  ,  7g,  T55,  T373 

2.  Make  a  decreasing  ratio  with  2s  and  2i,   with  12.45 
and  9f,     with  3  gal.  1  pt.  and  2  gal.  2  qt.,     with  13^  and 

1  K,5 


ART.  245.  Since  every  ratio  is  a  fraction  whose  numer- 
ator is  the  consequent,  and  denominator  the  antecedent, 
whatever  is  true  of  a  fraction  is  true  of  a  ratio;  hence, 

1st.  Multiplying  the  consequent  or  dividing  the  antecedent, 
multiplies  the  ratio.  (Arts.  112  and  115.) 

The  ratio  10  :  4  is  y*g  =  |;  if  the  consequent  be  multiplied  by  3, 
the  ratio  10  :  12  is  j-g  =f,  which  is  the  former  ratio  multiplied  by 
2;  if  the  antecedent  be  divided  by  2,  the  ratio  5  :  4  is  5,  which  is 
Ihc  former  ratio  multiplied  by  2. 

2d.  Multiplying  the  antecedent  or  dividing  the  consequent 
divides  the  ratio.  (Arts.  113  and  114.) 

REVIEW.  —  244.  How  many  ratios  can  be  formed  with  two  numbers? 
Give  an  example.  How  arc  they  distinguished?  What  is  a  ratio  of 
equality?  What  is  the  rule  for  making  an  increasing  or  decreasing  ratio! 
245.  How  is  a  ratio  multiplied?  Why?  How  is  a  ratio  divided  7  Why? 


PROPORTION.  191 


The  ratio  7  :  6  is  ^ ;  if  the  consequent  be  divided  by  3,  the  ratio 
7  :  2  is  7,  which  is  the  former  ratio  divided  by  3 ;  if  the  antecedent 
be  multiplied  by  2,  the  ratio  14  :  6  ia  -f$  =  |,  which  is  the  former 
ratio  divided  by  2. 

3d.  Multiplying  or  dividing  both  terms  of  a  ratio  by  a 
number,  does  not  alter  its  value.  (Arts.  116  and  117.) 

The  ratio  9:6  is  $  ==  f ;  if  both  terms  are  multiplied  by  2,  the 
ratio  18  :  12  is  jg  — f  8till>  if  botl1  terms  be  divided  by  3,  the 
ratio  6  :  4  is  g  =  |,  the  same  as  at  first. 


XIII.    PROPORTION. 

ART.  246.    Proportion  is  an  expression  of  equal  ratios. 

The  ratios  3  :  5  and  6  :  10  are  equal,  each  being  of  the 
value  3.  Placing  a  double  colon  (:  :)  between  them,  forms 
the  proportion  3  :  5  :  :  6  :  10,  read  3  is  to  5  as  6  is  to  10, 
or  the  ratio  of  3  to  6  is  equal  to  the  ratio  of  6  to  10. 

REMARK. — Either  ratio  may  be  written  first;  thus,  6  :  10  :  :  3  :  5 
is  the  same  as  3  :  6  :  :  6  :  10,  since  each  expresses  the  equality  of 
the  same  ratios. 

Instead  of  the  double  colon,  the  sign  of  equality  is  sometimes 
used;  as,  3  :  6  =  6  :  10,  is  the  same  as  3  :  6  :  :  6  :  .10. 

A  proportion  with  more  than  two  equal  ratios  is  called 
a  continued  proportion,  as3:5::6:10::-9:15;  but 
a  proportion  in  Arithmetic  generally  contains  only  two 
equal  ratios,  and  has  4  terms,  since  each  ratio  has  2  terms. 

Since  each  ratio  has  an  antecedent  and  consequent,  every 
proportion  has  two  antecedents  and  two  consequents,  the 
1st  and  3d  terms  being  the  antecedents,  and  the  2d  and 
4th  the  consequents.  - 

The  first  and  last  terms  of  a  proportion  arc  called  the 
extremes;  the  middle  terms,  the  means.  All  the  terms  are 
called  2)roportionalst  and  the  last  term  is  said  to  be  a  fourth 
proportional  to  the  other  3  in  their  order. 

Ratio  is  the  relation  between  two  numbers  shown  by  their  quotient: 
proportion  is  the  relation  between  two  ratios  shown  by  their  equality. 
The  former  has  two  terms,  the  latter  four. 


REVIEW. — 245.  IIow  is  a  ratio  altered  in  form  and  not  in  value? 
Why  ?  246.  What  is  Proportion  ?  Give  an  example.  How  is  it  written  7 
What  other  way?  What  is  a  continued  proportion  T 


92  HAY'S   HIGHER   ARITHMETIC. 


Three  numbers  are  in  proportion,  when  the  1st  has  the 
same  ratio  to  the  2d  as  the  2d  has  to  the  3d;  thus,  4,  8 
and  16  are  in  proportion,  for  4  :  8  :  :  8  :  It),  each  ratio 
being  2.  The  second  term  is  then  called  a  mean  propor- 
tional between  the  other  two;  and  the  last  term  a  third 
proportional  to  the  first  and  second. 

ART.  247.  Variation  is  a  general  method  of  expressing 
proportion  often  used,  and  is  either  direct  or  inverse. 

Direct  variation  exists  between  two  quantities  when  they 
increase  together,  or  decrease  together. 

Thus,  the  distance  a  ship  goes  at  a  uniform  rate,  varies  directly  as 
the  time  it  sails ;  which  means  that  the  ratio  of  any  two  distances 
Is  equal  to  the  ratio  of  the  corresponding  times  taken  in  the 
same  order. 

Inverse  variation  exists  between  two  quantities  when 
one  increases  as  the  other  decreases. 

Thus,  the  time  in  which  a  piece  of  work  will  be  done,  varies 
inversely  as  the  number  of  men  employed;  which  means  that  the 
ratio  of  any  two  times  is  equal  to  the  ratio  of  the  numbers  of  men 
employed  for  those  times,  taken  in  reverse  order. 

ART.  248.  Since  only  equal  ratios  form  a  proportion, 
to  determine  the  truth  of  a  proportion, 

Find  the  value  of  each  ratio  in  the  proportion;  if  these 
ratios  are  equal,  the  proportion  is  true  ;  if  not,  it  is  false. 

Thus,  8  :  10  :  :  12  :  15  is  a  true  proportion,  the  ratios  Jg°  and  '^ 
being  each  equal  to  | ;  but  9  :  5  :  :  8  :  2  is  not,  because  the  ratios 
f  and  f  are  not  equal,  the  former  being  |,  the  latter  $. 

WHICH  ARE  TRUE  PROPORTIONS,  AND  WHICH  NOT? 

1.  7  :  10  :  :  8  :  12,     and  4  :  3  :  :  24  :  18. 

2.  2  ft.  1  in.  :  1  yd.  4  in.  :  :  62^  ct.  :  §1. 

3.  16£  :  21  :  :  2  bu.  3  Pk.  :  3  yd.  2  qr. 

4.  3hr.l8min. :  fhr.SO min. :  :  2 gal.  Iqt. :  4gal.2qt. 


REVIEW. — 246.  How  many  ratios  in  a  proportion  generally?  How 
nany  antecedents?  How  many  consequents?  What  are  the  extremes? 
The  moans  ?  What  are  tho  terms  called  ?  The  last  term  ?  How  are  Ratio 
and  Proportion  distinguished  ?  When  are  three  numbers  in  Proportion  ? 
What  is  the  second  term  then  called  ?  What  is  the  third  term  called  ? 
247.  What  is  variation?  What  two  kinds?  What  is  direct  variation? 
Give  an  example.  What  is  inverse  variation  ?  Give  an  example. 


PROPORTION. 


5.  $5.331  :  $12  :  :  fcwt.  :  150  Ib. 

6.  76yd.  1ft.  lOin.  :  13rd.  7£ft.  :  :  $17.18}  :  $16| 

7.  16.208:943  ::9A.  1R.  6^P.  :  5  A.  3R.  37.64  P, 

8.  23  T.  5  cwt.  1  or.  15  Ib.  :  8  T.  2  cwt.  3  qr.  14  Ib.  :  : 

2  gal.  2  qt.  :  3  qt.  1  pt. 

9.  £3  17s.  8}d.  :  £8  10s.  3R  :  :  IGjbu.  :  35  bu.  3pk. 
10  22°12'4U":58°7'6.42"::1.3hr.:3hr.44min. 

11.  25  Ib.  Tr.  :  24  Ib.  av.  :  :  $6  :  $7. 

12.  5  yd.  2  ft.  9  in.  :  9E,F1.  3na.  :  :  $7.75 


PROPOSITION. 

ART.  249.  In  every  true  proportion,  the  product  of  the 
numbers  in  the  means  is  equal  to  the  product  of  those  in  the 
extremes. 

DEMONSTRATION  —  In  every  true  proportion,  as  6  :  3  :  :  10  :  6,  the 
ratios  are  equal,  viz  :  f  =  T6a.  If  both  terms  of  the  first  ratio  be 
multiplied  by  10,  and  both  terms  of  the  second  ratio  by  6,  their 
values  are  not  altered  and  they  are  still  equal,  viz:  f  £  [8  =T60>£  f 

The  denominators  of  these  fractions,  having  the  same  factors,  are 
equal;  to  make  the  fractions  equal,  the  numerators  must  also  be 
equal  ;  that  is,  3  X  10  =  6  X  6  ;  but  3  X  10  is  the  product  of  the 
numbers  in  the  means,  and  6X6,  the  product  of  those  in  the  ex- 
tremes ;  hence,  the  proposition  is  proved. 

COROLLARY  1.  —  Either  extreme  is  equal  to  the  product  of 
the  means  divided  by  the  other  extreme. 

COROLLARY  2.  —  Either  mean  is  equal  to  the  product  of 
the  extremes  divided  by  the  other  mean. 

The  truth  or  falsity  of  a  proportion  can  be  determined 
by  this  proposition  ;  the  corollaries  serve  to  find  any  term 
of  a  proportion,  when  the  other  three  are  known. 

Thus,  if  the  1st,  2d,  and  3d  terms  of  a  proportion  are  G,  10,  and 
15,  it  may  be  written  6  :  10  :  :  15  :  (  )  ;  the  4th  term  is  represented 
by  a  parenthesis,  and  found  by  Cor.  1  to  be  1  °  $  '  6  =  25.  The  un- 
known term,  25,  mast  be  of  the  same  denomination  as  the  other 
term,  15,  of  the  same  ratio. 

REMARK.  —  Before  applying  the  proposition  or  its  corollaries,  the 
Urrnt  of  each  ratio  must  be  of  the  same  denomination. 


REVIEW. — 248.    How  do  we  determine   the  truth   or   falsity  of  any 
proportion  ?    249.  What  relation  exists  between  the  extremes  and  meani 
in  every  true  proportion?    Prove  it 
17 


194  RAY'S  HIGHER  ARITHMETIC. 


FIND   THE   UNKNOWN   TERM   OP 

1.  12  :  16  :  :  (  )  :  5  and  3|  :  (  )  :  :  U  :  2. 

Ans.  84  and  5. 

2.  Ibu.  2pk.  6qt.  :  6bu.  3  pk.  :  :  $3.871  :  (  ). 

Ans.  $15.50 

3.  (  )  :  4hr.  30  min.  :  :  2£mi.  :  3. 375  mi. 

Ans.  3  hr.  20  min. 

4.  2.16 A..:  (  )  :  :  £13  6s.  8d.  :  £15.  Ans.  2. 43 A. 

5.  12  yd.  3  qr.  :  46  yd.  3  qr.  :  :  (  )  :  6  T.  1  cwt. 

Ans.  1  T.  13  cwt. 

6.  $162.56^  :  $270.931  :  :  234  men.  :  (  ). 

Ans.  390  men. 

7.  46°  31'  9"  :  (  )  :  :  3  hr.  36  min.  :  2  hr.  42  min. 

Ans.  34°  53'  21!" 

8.  $16  :  45  ct.  :  :  1  Ib.  9  oz.  av.  :  (  ).     Ana.  11}  dr. 

9.  5  mi.  3  fur.  30  rd.  :  7  mi.  10  rd.  ::(•):  54  horses 

Ans.  42  horses. 

10.     33  bu.  1  pk.  of   potatoes  :  (  )  :  :  4  bu.  U  pk.  of 
apples  :  2bu.  2  pk.  of  apples.       Ans.  19  bu.  of  potatoes. 


SIMPLE  PROPORTION. 

ART.  250.  Simple  Proportion  is  a  method  of  solving 
practical  questions  by  a  ratio  or  proportion ;  it  is  some- 
times called  the  Rule  of  Three,  because  the  answer  is 
obtained  by  finding  one  term  of  a  proportion  whose  other 
three  terms  are  known. 

If  5  qt.  of  strawberries  cost  75  ct.,  what  are  9  qt.  worth 
at  the  same  rate? 

SOLUTION  BY  ANALYSIS. — If  6qt.  cost  75  ct.,  Iqt.  costs  £  of  76  ct 
=  15  ct.,  and  9  qt.  cost  9  times  15  ct.  =  135  ct.  =  $1.35. 

SOLUTION  sr  FnaroRTiou. — Since  the  cost  of  each  quart  is  the 
same,  the  whole  cost  varies  directly  as  the  number  of  qt.;  that  is, 
9  qt.  being  §  of  5  qt.  are  worth  £  as  much,  or  §  of  75  ct.  =  9  ^7  6  ct 
=  $1.35.  Or,  the  ratio  of  the  quantities  being  the  same  as  the  ratio 
of  their  costs,  we  have  5  qt.  :  9  qt.  :  :  75  ct.  :  (  ),  in  which  the  re- 
quired term  is  found,  by  Cor.  Is*,  Art.  249,  to  be  9  V*  =  136  ct- 

REVIEW.— 249.  What  is  Cor.  1st?  Cor.  2d?  Of  what  use  is  the  pro 
position?  Of  what  uso  are  the  Corollaries  ?  Give  an  example. 


SIMPLE  PROPORTION.  195 

REMARK.  —  Sample  Proportion  is  sometimes  applied  to  questions 
which  do  not  admit  of  solution  in  that  w&y.  In  cases  of  doubt, 
resort  to  a  close  and  careful  analysis,  or  to  the  following  test  : 

A  question  to  be  solved   by   Simple   Proportion,  must 
contain   two  kinds  of  quantities,  and  two  of  each  kind 
three  of  the  quantities  must  be  known  and  one  required. 

The  two  given  quantities  which  are  of  different  kinds 
must  be  so  related,  that  when  one  is  doubled  the  other  ia 
necessarily  doubled  or  halved;  in  fact,  they  furnish  the 
rate  which  is  applied  to  the  remaining  given  quantity  to 
obtain  the  answer.  Hence, 

TO   SOLVE   QUESTIONS   BY   SIMPLE   PROPORTION, 

RULE.  —  With  the  two  given  quantities  which  are  of  the  same 
kind,  form  an  increasing  or  decreasing  ratio,  according  as  the 
answer  should  be  greater  or  less  than  the  third  given  quantity; 
multiply  the  third  quantity  by  this  ratio,  and  the  result  will  be 
the  answer  required. 

NOTE.  —  Express  the  ratio  in  its  simplest  form,  and  cancel  when 
possible. 

For  the  benefit  of  those  who  prefer  to  state  the  question 
in  the  form  of  a  proportion  before  commencing  the  opera- 
tion, we  give  the  following 

RULE.  —  Write  that  quantity  for  the  3d  term  which  is  of  the 
tame  denomination  as  the  answer;  next  to  it,  in  the  %d  term,  put 
the  larger  or  smaller  of  the  other  two  quantities,  according  as  the 
answer  should  be  larger  or  smaller  than  the  third  term.  The 
remaining  given  quantity  is  then  put  in  trie  first  term,  and  the 
fourth  or  required  term,  is  found  by  multiplying  the  second  and 
third  terms  together,  and  dividing  their  product  by  the  first. 

1.    If  2  1  Ib.   of  sugar  are   worth   7^  lb.  of  rice,  how 
much  sugar  is  19.81b.  of  rice  worth? 

19.8  19.8      11      217.8 

— 
30 


.  .  . 

SOLUTION.—  —  r  of  2f  lb.  =  —  r  X  —  =  —  —  =  7.26  lb.  Ant. 
7£  7j        4 


19.8X2-1 
Or,  7-|  :  19.8  :  :  2|  :  (  );  4th  term-=  -  ri  —  i=  7.26  lb. 

7*  f 


REVIEW.  —  249.  Of  what  denomination  will  the  required  term  be? 
Before  applying  the.  proposition  or  corollaries,  what  should  he  done? 
250.  What  is  Simple  Proportion?  What  is  it  sometimes  called?  Why? 
Give  an  example.  What  is  the  solution  by  analysis?  By  proportion! 
What  kind  of  questions  properly  come  under  simple  proportion  ? 


196  RAY'S   HIGHER   ARITHMETIC. 


2.  If  15  men  do  a  piece  of  work  in  9|da.,  how  long 
will  36  men  be  in  doing  the  same? 

SOLUTION. — Since  36  men  will  require  lett  time  than  15  men  to 
do  the  same  work,  the  answer  should  be  less  than  9|da. ;  make  a 
decreasing  ratio,  |  g,  and  multiply  the  remaining  quantity  by  it: 

if  xaf-if  xV  «=*<**•  Ans- 

3.  If  I  walk  lOAmi.  in    3  hr.,  how  far  will  I  go  in 
10  hr.,  at  the  same  rate?  Ans.  35  mi. 

4.  If  the  fore-wheel  of  a  carriage  is  8ft.  2  in.  in  cir- 
cumference, and    turns    round  670    times,  how  often  will 
the  hind-wheel,  which  is  11  ft.  8'in.  in  circumference,  turn 
round  in  going  the  same  distance?          Ans.  469  times. 

5.  If  a  horse  trot  3  mi.  in  8  min.  15  sec.,  how  far  can 
he  trot  in  an  hour  at  the  same  rate?          Ans.  21i9T  mi- 

6.  What  is  a  servant's  wages  for  3  wk.  5  da.,  at  $1.75 
per  week?  Ans.  $6.50 

7.  What  is  a  bbl.  of  powder  containing  132  Ib.  worth, 
if  15  Ib.  are  sold  for  $5.43f  ?  Ans.  $47.85 

8.  A  body  of  soldiers  are  42  in  rank  when  they  are 
24  in   file:    if  they  were   36   in   rank,  how  many  in   file 
would  there  be  ?  Ans.  28. 

9.  If  a  pulse  beats   28  times   in   16  sec.,  how  many 
times  a  minute  is  that?  Ans.  105. 

10.  On  15  successive  squares  are  614  houses:  how  far 
from  No.  1  is  No.  277?    Ans.  6^14,  or  nearly  7  squares. 

11.  If    a   cane   3ft.    4  in.  long,   held    upright,    casts  a 
shadow  2  ft.  1  in.  long,  how  high  is  a  tree  whose  shadow 
at  the  same  time  is  25  ft.  9  in.?  Ans.  41  ft.  2|  in. 

12.  If  a  horse  draw  25  bu.   of  coal,  each  80  Ib.,   how 
many  bu.  of  coke,  each  96  Ib.,  can  he  draw?     Ans.  20i 

13.  If  a   farm  of  160  A.  rents    for  $450,   how  much 
should  be  charged  for  one  of  840  A.?     Ans.  $2362.50 

14.  If  18d.  sterling,  equals  25  ct.  U.  S.  money,  what 
s  a  half-crown  (2s.  6d.)  worth?  Ans.  4l3ct. 

15.  A  grocer  has  a  false  gallon,  containing  3  qt.  lA  pt. : 
what  is  the  worth  of  the   liquor    that   he  sells  for   $240, 
*nd  his  gain  by  the  cheat?    'Ans.  $225,  and  $15  gain. 

16.  If  he  uses  14|  oz.  for  a  pound,  how  much  docs  he 
cheat  by  selling  sugar  for  $27.52?  AM.  $2.15 


REVIEW. — 250.  What  is  the  rulo  for  solving  questions  by  Simple  Pro- 
portion? How  should  the  ratio  be  expressed?  What  is  the  ordinary  rul« 
for  simple  proportion  T 


SIMPLE   PROPORTION. 


17.  An  equatorial  degree  is  365000  ft.:  how  many   ft 
in  80°  24'  37"  of  the  same?  Am.  29340751 75 

18.  If  a  pendulum  beats  5000  times  a  day,  how  often 
does  it  beat  in  2hr.  20  min.  5  sec.?      -4ns.  486]  I \  times. 

19.  If  I  do  a  piece  of  work  in  108  days,  of  83  hr., 
how  many  days  of  6-4  hr.  would  I  be?  -4ns.  136. 

20.  A  man  borrows  $1750,  and  keeps  it  1  yr.  8mon.: 
how  long  should  he   lend  $1200  to  compensate  for  the 
favor?  Ans.  2  yr.  5  mon.  5  da. 

21.  A  garrison  has  food  to   last   9  mon.,  giving  each 
man  1  Ib.  2  oz.  a  day:  what  should  be  a  man's  daily  allow- 
ance, to  make  the  same  food  last  lyr.  8mon.?  Ans.  SAoz. 

22.  A  garrison  of  560    men    have   provisions   to    last 
during  a  siege  at  the  rate  of  lib.  4  oz.  a  day  per  man;  if 
the  daily  allowance  is  reduced  to  14  oz.  per  man,  how  large 
a  reinforcement  could  be  received?  Ans.  240  men. 

23.  A  shadow  of  a   cloud   moves  400ft.  in    18.f  sec. : 
what  was  the  wind's  velocity  per  hour?      -4ns.  14/1  nii- 

24.  If  1  Ib.  Troy  of  English  standard  silver  be  worth 
£3  6s.,  what  is  1  Ib.  av.  worth?  Ans.  £4  2R 

25.  If  I  go  a  journey  in  12|  days,  at  40  mi.  a  day,  how 
long  would  I  be  at  29 1  mi.  a  day?  Ans.  17-r  da. 

26.  If  f  of  a  ship  are  worth  $6000,  what  is  the  whole 
of  it  worth?  Ans.  $10800. 

SUGGESTION. — The  two  similar  quantities  in  this  question,  are 
5  ninths  of  the  ship,  and  9  ninths  of  the  ship. 

27.  If  A,  worth   $5840,  is  taxed   $78.14,  what  is  B 
worth,  who  is  taxed  $256.01?  Ans.  $19133.59 

28.  What  are  4  Ib.  6  oz.  of  butter  worth,  at  2 Set.  a  Ib.? 

-4ns.  $1.224 

29.  If  I  gain  $160.29  in  2  yr.  3  mon.,  what  would  I 
gain  in  5  yr.  6  mon.,  at  that  rate?  Ans.  $391.82 

30.  If  I  gain  $92.54  on  $1156.75  worth  of  sugar,  how 
much  must  I  sell  to  gain  $67.32?  -4«s.  $841.50  worth. 

31.  If  coffee  costing  $255  is  now  worth  $318.75,  what 
did  $1285.20  worth  cost?  Ans.  $1028.16 

32.  If  I  gain  $7.75  by  trading  with  $100,  how  much 
ought  I  gain  on  $847.56?  -4ns.  $65.6859 

33.  A  has  cloth  at  $3.25  a  yd.,  and  B   has  flour  at 
$5.50  a  bbl.     If,  in  trading,  A  puts  his  cloth  at  $3.022, 
what  should  B  charge  for  his  flour?  -4ns.  $6.13/3 

34.  What  is  a  pile  of  wood,  15ft.  long,  10A  ft.  high, 
and  12ft.  wide,  worth,  at  $4.25  a  cord?    Ans.  $62.75 


198  RAY'S   HIGHER   ARITHMETIC. 


35.  Find  7  mon.  rent,  at  $4*75  a  yr.     Ans.  $277-08 

36.  If  a  boat  is  rowed  at  the  rate  of  6  miles  an  hour, 
and  is  driven  44  feet  in  9  strokes  of  the  oar,  how  many 
strokes  are  made  in  a  minute?  Ans.  108  strokes. 

37.  If  g  of  a  yd.  of  cloth  cost  $3£,  what  is  the  worth 
of  I  of  an  E.  E.?  Ans.  $13^ 

In  Fahrenheit's  thermometer,  the  freezing  point  of  water 
is  marked  32°,  and  the  boiling  point  212°  :  in  the  Centi- 
grade, the  freezing  point  is  0°,  and  the  boiling  point  100°: 
in  Reaumer's,  the  freezing  point  is  0°,  and  the  boiling 
point  80°. 

38.  From  these  data,  find  the  value  of  a  degree  of  each 
thermometer  in  the  derees  of  the  other  two.     Ans.  1°  F. 


=  2|°F. 

39.  Convert  108°  F.  to  degrees  of  the  other  two  ther- 
mometers.   (First  subtract  32°.)     Ans.  33?°  R.  and  42'^°  C. 

40.  Convert  25°  R.  to  degrees  of  the  other  two  ther- 
mometers. Ans.  31i-°C.  and  881°  F. 

41.  Convert  46°  C.  to  degrees  of  the  other  two  ther- 
mometers. Ans.  36f°R.  and  114*°  F. 

.  In  the  working  of  machinery,  it  is  ascertained  that  the 
available  power  is  to  the  weight  overcome,  inversely  as  the 
distances  they  pass  over  in  the  same  time. 

42.  If  the  whole  power  applied  is  180  Ib.  and  moves 
4ft.,  Tiow  far  will  it  lift  a  weight  of  960  Ib.?      Ans.  6  in. 

NOTE.  —  The  available  power  is  taken  -f  of  the  whole  power,  ^  being 
allowed  for  friction  and  other  impediments. 

43.  If  512  Ib.  bo  lifted  1  ft.  3  in.  by  a  power  moving 
6  ft.  8  in.,  what  is  the  power  ?  Ans.  144  Ib. 

SUGGESTION.  —  Find  the  available  power;  then  add  |  of  itself. 

44.  A  lifts  a  weight  of  1440  Ib.  by  a  wheel  and  axle; 
for  every  3  ft.  of  rope  that  passes  through  his  hands,  the 
weight  rises  42in:  what  power  does  he  exert?  Ans.  270  Ib. 

45.  A  man  weighing  198  Ib.  let  himself  down    54  ft, 
with  a  uniform  motion,  by  a  wheel  and  axle  :  if  the  weight 
at  the  hook  rises  12  ft.,  how  much  is  it?        Ans.  594  Ib. 

46.  Two  bodies  free   to   move,  attract  each  other  with 
forces  that  vary  inversely  as  their  weights.     If  the  weights 
are  9  Ib.  and  4  Ib.  and  the  smaller  is  attracted  10  ft.,  how 
far  will  the  larger  be  attracted?  Ans.  4  ft.  5?  in. 

47.  Suppose  the  earth  and  moon  to  approach  each  other 
in  obedience  to  this  law,  their  weights  being  49147  and 


COMPOUND   PROPORTION.  199 

123  respectively.     How  many  miles  would  the  inoon  move 
while  the  earth  moved  250  miles?         Am.  99892+  mi. 

Can  the  following  questions  be  solved  by  proportion  ? 

(See  Rem.  Art.  250.) 

48.  If  3  men  mow  5  A.  of  grass  in  a  day,  how  many 
men  will  mow  13^  A.  in  a  day? 

49.  If  6  men  build  a  wall  in  7  da.,  how  long  would  10 
men  be  in  doing  the  same  ? 

50.  If  I  gain  15  cents  each,  by  selling  books  at  $4.80 
a  doz.,  what  is  my  gain  on  each  at  $5.40  a  doz.? 


COMPOUND  PROPORTION. 

ART.  251.  Compound  Proportion  is  a  method  of  solving 
questions  by  the  use  of  a  compound  ratio,  or  by  more  than 
one  proportion. 

The  questions  to  which  Compound  Proportion  is  applied 
resemble  those  under  simple  proportion ;  but  the  value  of 
the  quantity  required  depends  not  on  one  pair,  but  on  two 
or  more  pairs  of  similar  quantities :  for  instance, 

If  3  men  mow  8  A.  of  grass  in  4  da.,  how  long  would 
10  men  be  in  mowing  36  A.? 

SOLUTION. — If  the  10  men     1st  step.     T30  of  4  da. 
were  to  mow  8  A.  instead  of      2d  step.       W  of  A  of  4  d». 
36  A.,  the  question  would  be       _r2j        * 

/•    rt-  i         T»  i«  1  —~    "8»   **«*•      AllS. 

one  of  Simple  Proportion,  and 

the  answer  would  be  -^j  of  4  da.,  as  shown  in  the  1st  step  of  the 
operation ;  but  this  result  being  the  time  in  which  10  men  mow  8  A, 
the  time  in  which  they  will  mow  36  A.  will  be  3g6  of  it=  3g6  of  ^ 
of  4  da.,  as  shown  in  the  2d^  step,  which  reduces,  by  cancellation, 
to  6f  da.  Hence,  a  question  in  Compound  Proportion,  is  only  a  suc- 
cession of  questions  in  Simple  Proportion,  each  of  which  gives  a  result 
to  be  used  in  the  next,  and  the  last  result  is  the  answer  required. 

After  sufficient  practice,  the  successive  steps  may  be 
dispensed  with,  and  the  answer  written  as  a  compound 
fraction  at  once. 

TO   SOLVE   QUESTIONS   BY  COMPOUND   PROPORTION, 

RULE. — Form  an  increasing  or  decreasing  ratio  of  each  pair 
of  similar  quantities,  as  if  the  answer  depended  only  on  (host 


200  RAY'S  HIGHER   ARITHMETIC. 


two  and  the  odd  term;    multiply  these  ratios  and  the  odd  term 
together;  the  product  will  be  the  answer, 

NOTE. — The  odd  term  is  the  one  which  is  unlike  all  the  rest 

To  TEACHEBS. — Pupils  should  analyze  each  example  thoroughly, 
and  give  the  reasons  for  every  step :  such  a  course  will  be  a  valu- 
able training  to  the  mental  powers,  and  is  the  only  way  to  clear  up 
an  otherwise  obscure  and  difficult  subject. 

If  the  proportional  form  is  preferred,  use  a  succession 
of  simple  proportions,  as  already  explained,  or  the  rule 
in  "  Ray's  Arithmetic,"  3d  Book,  Art.  205.  A  simple 
mode  of  stating  questions  in  proportion  is  this 

RULE   OF  CAUSE  AND   EFFECT. 

Separate  all  the  quantities  contained  in  the  question  into  two 
causes  and  their  effects. 

Write,  for  the  first  term  of  a  proportion,  all  the  quantities 
that  constitute  the  first  cause ;  for  the  second  term,  all  that  con- 
stitute the  second  cause ;  for  the  third,  all  that  constitute  the  effect 
of  the  first  cause;  and  for  the  fourth,  all  that  constitute  the 
effect  of  the  second  cause.  The  required  quantity  may  be  indi- 
cated by  a  bracket,  and  found  by  one  of  the  rules  in  Art.  249. 

NOTE. — The  two  causes  must  be  exactly  alike  in  the  number  and 
kind  of  their  terms  ;  and  so  must  the  two  effects. 

1.  If  6  men,  in  10  days  of  9  hr.  each,  build  25  rd.  of 
fence,  how  many  hours  a  day  must  8  men  work  to  build 
48  rd.  in  12  days? 

SOLUTION. —  6  men  10  da.  and  9  hr.   constitute  the  1st  catise, 
whose  effect  is  25  rd. ;  8  men  12  da.  and  (  )  hr.  constitute  the  2d 
cause,  whose  effect  is  48  rd.     Hence, 
6  men.        8  men. 

10  da.     :      12  da.   :    :   25  rd.   :   48  rd. 
9hr.  (  )hr. 

And  the  required  term  is  6osx8X**8  =  10|  hr.  (Art  249,  Cor.  2.) 

2.  A  boy  makes  8  steps,  each  1ft.  10  in.,  \rhile  a  man 
makes  5  steps,  each    2ft.  8  in.:  how  far  will  the  boy  go 
while  the  man  is  going  3|  miles?  Ans.  4g  mi. 

REVIEW. — 251.  What  is  Compound  Proportion  ?  What  does  the  valne 
of  the  quantity  required  depend  on  ?  Solve  the  example.  What  is  tba 
rule?  What  is  the  odd  term?  What  is  the  rule  of  cause  and  effect? 
What  is  said  of  the  two  causes  ?  The  two  effects  ? 


COMPOUND  PROPORTION.  201 

3.  If  18  pipes,  each  delivering  6  gal.  per  minute,  fill 
ft  cistern  in  2  hr.  16min.,  how  many  pipes,  each  delivering 
20  gal.  per  minute,  will  fill  a  cistern  7£  times  as  large  as 
the  first,  in  3  hr.  24min.?  Ans.  27. 

4.  The  use  of  $100  for  1  year  is  worth  §8:  what  Is 
the  use  of  $4500  for  2yr.  8mon.  worth?      Ail's.  $900. 

5.  If  12  men  mow  25  A.  of  grass  in  2  da.  of  lOjhr., 
how  many  hours  a  day  must  14  men  work   to  mow  ,-ID 
80  A.  field  in  C  days?  Am.  9|hr. 

6.  If  4  horses  draw  a  railroad  car  9  miles  an   hour 
how  many  miles  an  hour  can  a  steam  engine  of  150  horse- 
power drive  a  train  of  12  such  cars,  the  locomotive  and 
tender  hcing  counted  3  cars?  Ans.  22^  nii.  per  hr. 

7.  How  many  half  eagles,  each  weighing  5  pwt.  9  gr. 
and  made  of  gold  79a  pure,  are  equivalent  to  1000  Eng- 
lish sovereigns  each  weighing  5  pwt.  3.274gr.,  and  made 
of  gold  H  pure?  Ant.  973^1:1 

8.  If  the  use  of  $3750  for  8  mon.  is  worth  $68.75, 
what   sum   is    that  whose  use   for   2  yr.   4  mon.   is  worth 
$250?  Ans.  $3896.10-f. 

9.  If  the   use   of  $1500  for  3  yr.  8  mon.    25  da.   is 
worth  $336.25,  what  is  the  use  of  $100  for  1  yr.  worth? 

Ans.  $6. 

10.  If   240  panes  of   glass    18  in.   long,    10  in.  wide, 
glaze   a   house,  how   many   panes   16  in.    long   by   12  in. 
wide  will  glaze  a  row  of  6  such  houses?         Aits.  1350. 

11.  If  it  require  800  reams  of  paper  to  publish  5000 
volumes  of  a  duodecimo  book  containing  320  pages,  how 
many  reams  will  be  needed  to  publish  24000  copies  of  a 
book,  octavo  size,  of  550  pages?          Ans.  9900  reams. 

12.  A  man  has  a  bin  7  ft.  long  by  £5  ft.  wide,  and  2ft. 
deep,  which  contains  28  bu.  of  corn:  how  deep   must  he 
make   another,  which  is   to  bo  18ft.  long  by  lg  ft.  wide," 
in  order  to  contain  120bu.?  -4ns.  4g  ft. 

13.  If  it  require  4500  bricks,  8  in.  long  by  4  in.  wide, 
to  pave  a  court-yard  40  ft.  ?ong  by  25  ft.  wide,  how  many 
tiles  10  in.  square,  will  be  needed   to  pave   a  hall  75ft, 
long  by  16  ft.  wide?  Ans.  17-28. 

14.  If  150000  bricks  are  used  for  a  house  whose  walls 
average   lift,   thick,   30ft.    high,  and    210ft.  long,  how 
many  will  build  one  with  walls  2ft.  thick,  24ft.  hitrh,  and 
824ft.  long?  Au».  240000. 

15.  A   garrison    of   1800  men   has    provisions   to   last 
£2  mon.  at  the  rate  of  1  Ib.  4  oz.  a  day  to  each;  how  long 


202  RAY'S   HIGHER  ARITHMETIC. 

will  5  times  as  much  last  3500  men,  at  the  rate  of  12  oz 
per  day  to  each  man  ?  Am.  1  yr.  7?  moil. 

16.  What  sum  of  money  is  that  whose  use  for  3  yr.,  at 
the  rate  of  $44  for  every  hundred,  is  worth  as  much  as 
the  use  of  $540  for  1  yr.  8  mon.,  at  the  rate  of  $7  for 
every  hundred?  Ans.  $466. 66§ 


XIV.  PERCENTAGE. 

ART.  252.  PERCENTAGE  is  derived  from  the  Latin 
phrase  per  centum,  meaning  on  the  hundred.  It  is  under- 
stood to  embrace  all  those  operations  in  which  reference  is 
made  to  100  as  a  unit  of  comparison. 

Its  applications  are,  Gain  and  Loss,  Commission,  In- 
terest, Banking,  Stocks,  Insurance,  Duties  and  Taxes. 

If  I  have  $750  in  business,  and  gain  $180,  what  is  my 
gain  on  every  $100? 

SOL DT JON. — This  is  a  question  in  Simple  Proportion,  and  is 
solved  as  follows :  $760  :  $100  :  :  $180 :  (  );  the  4th  term  is  found 
by  Cor.  1st,  Art.  249,  to  be 

100  X  180 


750 


=  $24.  Am. 


Hence,  I  gain  $24  on  every  $100,  or  24  per  cent.,  since  per  cent. 
means  on  the  hundred. 

The  $24,  in  the  solution  above,  is  called  the  rate  per 
cent,  of  the  $180  to  the  $750. 

Though  rate  per  cent,  strictly  means  the  number  on  a 
100,  yet  as  "$24  on  a  $100"  conveys  the  same  idea  as 
"24  hundredths,"  and  the  latter  is  simpler  and  more 
general,  it  is  the  prevailing  practice  to  consider  the  rate 
per  cent,  that  one  number  is  of  another,  as  the  number  of 
hundredths  it  is  of  that  other,  and  any  per  cent,  of  a  num- 
ber is  so  many  hundredths  of  it. 

Thus,  7  per  cent,  of  any  thing  is  7  hundredths  of  it. 

2^     do is  2^      do.       do. 

100  do is  100     do.       do.    or  the  whole. 

REVIEW. — 252.  What  is  the  meaning  of  Percentage?  What  does  it 
embrace  ?  What  are  its  principal  applications  ?  Solve  the  example.  What 
does  rate  percent,  mean  strictly? 


PERCENTAGE.  203 


'2.  If  I  have  160  sheep,  and  sell  35  per  cent,  of  them 
how  many  sheep  do  I  sell? 

SOLUTION. — The  whole  flock  (160)  being  100  per  cent.,  the  pro- 
portion is,  160  sheep  :  (  )  sheep  :  :  100  per  cent. :  85  per  cent.;  the 
required  term  is  '  6i°0^3  5  =  ^  sheep.  Ans. 

3,    In  a  battle,  78  men   are  killed,  which  is  13i  per 
cent,  of  the  whole  force;  how  many  were  engaged? 

SOLUTION. — The  whole  force  rs  100  per  cent.;  the  slain  (78)  are 
183  per  cent.;  hence,  (  )  men  :  78  men  :  :  100  per  cent. :  13^  per 
cent. ;  the  required  term  is 

78  X 100      3X78X100      ,R, 
18T"          ~40~  l5men'     ^ 

Hence,  there  are  three  cases  of  percentage,  according  as 
the  first  of  the  two  numbers  compared,  the  second,  or 
their  rate  per  cent.,  is  to  be  found.  All  these  cases  can 
be  solved  by  the 

GENERAL  RULE  FOR  PERCENTAGE. 

Any  two  numbers  are  to  each  other,  as  the  rates  per  cent,  they 
represent  of  the  same  quantity. 

NOTE. — If  one  of  the  numbers  is  the  standard  of  comparison 
for  the  other,  it  will  be  100  per  cent.;  if  they  are  both  referred  to 
a  third  quantity,  it  will  be  100  per  cent.,  and  their  rates  per  cent, 
will  depend  upon  the  nature  of  their  relations  to  it. 

Although  this  rule  covers  the  whole  subject,  it  is  too 
general  for  practical  purposes;  therefore  a  special  rule 
will  bo  given  for  each  case. 

CASE   I. 

TO   FIND   ANY   GIVEN    PER   CENT.   OP   A   GIVEN  NUMBER, 

ART.  253.  RULE. — Multiply  the  given  number  by  the  given 
rate  per  cent.,  and  divide  *he  product  by  100. 

NOTES. — 1.  It  often  saves  work  to  indicate  and  cancel. 

2.  The  dividing  by  100  may  generally  be  done  by  pointing  off  in 
the  product  two  more  decimal  places  than  are  in  the  multiplicand  ana 
multiplier. 

Find  9  percent,  of  $182.50  Si  82  50 

SOLUTION. — 9  per  cent,  of  $182.60  =  y§-g  of  Q 

$182.50  =  9  times  $182.50,  divided   by  100,  as  ^ 

the  rule  directs.  $16.4250 


204  RAY'S  HIGHER  ARITHMETIC. 

REMARK. — We  may  divide  by  100  first,  and  then  multiply  by 
the  rate  per  cent.;  or,  multiply  by  the  rate  per  cent,  written  ai 
decimal  hundredths. 

1.  Find  6  per  cent,  of  $82.372  Ans.  $4.94J 

2.  14s  pr.  ct.  of  6yd.  2ft.  Sin.   Ans.  2  ft.  11.96  in. 

3.  42  per  cent,  of  $1250.  Ans.  $525. 
REMARK. — Business  men  use  instead  of  the  words  "per  cent 

the  character  %  :  thus  42  per  cent,  is  written  42  fe. 

4.  Find  62^  %  of  1664  men.          AM.  1040  men. 

5.  35  %  of  * .  f  X  rife  =  77o°o  =  T>O  ^«- 

6.  ISO%  of4|  AM.  718 

7.  98  %  of  14  cwt.  2  qr.  20  Ib. 

=  14  cwt.  Iqr.  15!  Ib. 

8.  9|  %  of  48  mi.  6  fur.  16  rd.  =  4  mi.  4  fur.  24  rd. 

9.  33i  %  of  127  gal.  3  qt.  1  pt.  =  42  gal.  2  qt.  1  pt. 

QQ-i 

SOLUTION. — Since  -—  =  |,  take  \  of  the  given  number. 

10.  Find  1U  %  of  $3283.47.  Ans.  $364.83 

11.  40^  of  6hr.  28min.  15  sec. 

Ans.  2  hr.  35  min.  18  sec. 

12.  675^  of  31b.  10  oz.  16pwt.  22  gr. 

Ans.  26  Ib.  4  oz.  4pwt.  4jgr. 

13.  104  %  of  75  A.  1 R.  35 P.     =78  A.  IE.  38 P. 

14.  15  f  %  of  a  book  of  576  pages.      Ans.  90  pages. 

15.  23i%of45ct.  ^Hs.lOlct. 

16.  How  much  is  18|  %  of  a  cargo  that  weighs  41 6  T. 
15  cwt.  20  Ib.?  ^/i8.  78  T.  2  cwt.  3  qr.  10  Ib. 

17.  Find  337^  %  of  11  AM.  4£ 

18.  561  %  of  144  cattle.  Ans.  81  cattle. 

19.  16|  %  of  1932  hogs.  Ans.  322  hogs. 

20.  871  %  of  1634  C.  72  cu.  ft.  AM.  1430  C.  31  cu.ft. 

21.  1000^  of  $5.431  AM.  $54.37i 

22.  61  %  of  i?  AM.  fr 

23.  2j  #ofj AM.!, 

REVIEW. — 252.  What  is  the  rate  per  cent  that  one  number  ia  of  anotbrt 
considered?  What  is  any  per  cent,  of  a  number  ?  Give  examples.  Sol7fl 
example  2.  Also,  example  3.  How  many  cases  of  percentage?  Whut 
are  they?  What  is  the  general  rule  for  percentage?  If  one  of  the  given 
quantities  is  the  standard  of  comparison  for  the  other,  what  rate  per  cent 
•rill  it  represent  ? 


PERCENTAGE.  205 


24     What  part  is  25  %  of  a  farm  ?  Ans.  T2055  =  \ 

25.  What  part  of  a  quantity  is  6]  %  of  it?   also  8|  %; 
10#;   12£#;   16i#;  20%;   33*  £;   50%;  661%; 
and  75  %  ?        4«,.  T'g,  T'2,  T'5,  $,  J,  £,  J,  £,  f ,  |  of  it. 

26.  What  part  of  a  quantity  is  18|  %  of  it;  31J  %] 
37*%;  432%;  56j%;  621%;  68}  %  ;  81-j  % ;  83i  %; 
87i%;  and  931%? 

Ans.  T3s>  T6s>  i>  T7e»  T9g>  i  lii  ii»  l>  I»  is  of  it. 

27.  How  much  is  100  %  of  a  quantity;  125  %  of  it; 
250^;    675^;    1000%;    9437i%? 

Ans.  1  time,  IT  times,   2i,    6|,    10,    94|  times  it. 

28.  A  man  owning  |  of  a  ship,  sold  40  %  of  his  share: 
what  part  of  the  ship  did  he  sell,  and  what  part  did  he 
still  own?  Ans.  •&  sold;  ?9o  left. 

29.  A  owed  B  a  sun:  of  money;  at  one  time  he  paid 
him  40  %  of  it;  afterward  he  paid  him  25  %  of  what  he 
owed ;   and    finally  he    paid    him  20  %   of  what  he  then 
owed:  how  much  does  he  still  owe?  Ans.  -£5  of  it. 

30.  What  is  78£  %  of  12  T.  6cwt.  81b.? 

Ans.  9  T.  12  cwt.  1  qr. 

o31.    Out  of  a  cask  containing  47  gal.  2  qt.  1  pt.,  leaked 
6|  %'.  how  much  was  that?  Ans.  3  gal.  l§pt. 

32.  A  has  an  income  of  $1200  a  year;  he  pays  13  % 
of  it   for   board;    lOf  %  for   clothing;  6|  %  for  books; 
T\  %   f°r   newspapers;    12g  %  for  other  expenses:    how 
much  does  he  pay  for  each  item,  and  how  much  does  he 
save  at  the  end  of  the  year? 

Ans.  $156,  board;  $124.80,  clothing;  $81,  books;  $7, 
newspapers;  $154.50,  other  expenses;  $676.70,  saved. 

33.  Find  |   %  of  $1200.  Ans.  $10. 

34.  £    %  of  $47. 75.  Ans.  231  ct. 

35.  10  %  of  20  %  of  $13.50.  Aiu.  27  ct. 

36.  40  %  of  15  %  of  75  %  of  $133.331.  Ans.  $6. 

37.  8^  of  62i%  of  150^  of  $462.50.^ns.  $34.681 

38.  A   man    contracts   to   supply  dressed   stone   for  a 
court-house   for  $119449,   if  the  rough   stone  costs  him 


R  B  v  I  B  w.— 253.    What  is  Case  1  ?     The  rule  ?     How  is  the  division 
by  100  performed  ?     How  may  the  operation  be  performed  T 


206  RAY'S   HIGHER   ARITHMETIC. 

16  ct.  a  cu.  ft. ;  but  if  he  can  get  it  for  15  ct.  a  cu.  ft., 
he  will  deduct  3  %  from  his  bill;  how  many  cu.  ft.  would 
be  needed,  and  what  does  he  charge  for  dressing  a  cu.  ft.? 
Ans.  358347  cu.  ft.,  and  I7i  ct.  a  cu.  ft. 

39.  48  cfo  of  brandy  is  alcohol;  how  much  alcohol  does 
a  man  swallow  in  40  years  if  he  drinks  a  gill  of  brandy 

times  a  day?  Ans.  657gal.  1  qt.  1  pt.  2. 4  gills. 

CASE  II. 

ART.  254.  Two  numbers  being  given,  to  find  the  rate 
per  cent,  one  is  of  the  other. 

RULE. — Multiply  the  number  which  is  to  be  the  rate  per  cent, 
by  100,  and  divide  the  product  by  the  other  number. 

Or,  when  the  numbers  are  small,  Take  such  a  part  of  100 
os  the  number  which  is  to  be  the  rate  per  cent,  is  of  the  other. 

PROOF. — With  the  rate  per  cent,  thus  obtained,  and 
the  number  which  is  the  standard  of  comparison,  proceed 
by  the  last  rule;  if  the  result  is  the  same  as  the  other 
number,  the  work  is  right. 

18  is  how  many  percent,  of  276? 

SOLUTION. — 18  is  how  many  per  cent,      9  76^)1  8  0  0  f6A? 
of  276,  means  18  is  how  many  hundredths 
of  276.     Now  1  hundredth  of  276  is  f  g§, 

and  as  often  as  18  contains  this,  so  many  ..  o\144        12 

hundredths  will  it  be  of  276;  but  18-=-  /276  ~ 

fft  =  If  X  tfg,   (Art.   131),  =  1800  -      ^   g,  |         cent> 
276,  as  the  rule  directs. 

6  is  how  many  per  cent,  of  9? 

SOLUTION. — 6  is  ^  or  |  of  9,  and  to  find  how  many  hundredtLs  of 
0  it  is,  convert  f  into  hundredths.  To  do  this,  say  f  =  f  of  1  unit 
=  |  of  100  hundredths  =  66|  hundredths  =  6G§  %\  or,  multiply  both 
terms  of  f  by  a  number  that  will  make  the  denominator  100 ;  here, 

f\(*  ~ 

the  multiplier  is  33^,  which  changes  |  to  — ^  =  66§  %  as  before. 

1.  $14.40  is  how  many  %  of  $54?    Ans.  26|  %. 
SUGGESTION. — Reduce  both  to  ct. ;  and  generally  reduce  com- 
pound numbers  to  the  same  denomination  before  applying  tue  rule. 

2.  9  is  how  many  %  of  6?  Ans.  150  %. 

KB  VIEW. — 254.  What  is  Case  2d  ?  The  rule?  When  the  numbara  ar« 
mall  what  rule  can  be  used?  Explain  example  1.  Example  2. 


PERCENTAGE.  207 


3.  15  ct.  is  how  many  %  of  $2?  Ant.  7  2 

4.  2yd.  2ft.  Sin.  is  how  many  %  of  4rd.?    Am.  12^ 

5.  3  gal.  3qt.  is  what  %  of  31.2  gal.?       An*.lltf 

6.  |  is  how  many  %  of  $?  .4ns.  83  £ 
SOLUTION.  —  f  =  {§  and  |  =  }f  ;    the  first  is   !-£  01   |  of  tht 

last,  but  §  =  |  of  100  %  =  83|  %. 

7.  |  is  how  many  %  of  |?  Ans.  250. 

8.  2  of  |  of  4  is  what  %  of  1TV  4n*.  lOiYs 

9.  2£  is  how  many          %  of  3|?  .Ans.  663 

10.  |    is  how  many          %  of  H?  ^«s-  8633 

11.  ^  is  how  many         $  of  ^?  Am.  222| 

12.  $5.12  is  what  %  of  $640?  Am.  f 

13.  14|  is  how  many      %  of  2|?  Ans.  5172 

14.  $3.20  is  what    '       %  of  $2000?  Ans.  Ws 

15.  $45  is  what  %  of  $12?  ^TW.  375 

16.  750  men  is  what      %  of  12000?  Ans.  63 

17.  8|  is  how  many         %  of  I?  ^TW.  1050. 

18.  $7.29  is  what          %  of  $216?  Ans.  3| 

19.  $  is  how  many  %  of  §?  -4«».  192§   • 

20.  3qt.  I2pt.  is  what    %  of  5  gal.  23qt.?  Am.  16| 

21.  16bu.  32pk  is  what  %  of  7.125bu.?  Ans.  236JI 

22.  A's    money   is   50  %  more    than    B's;    then    B's 
money  is  how  many  %  less  than  A's?  Ans.  33  3 

SOLUTION.  —  Call  B's  money  100  %\  A's  money  is  then  160^: 
their  difference  is  60  <&,  which,  compared  with  A's  money,  150  % 


23.  If  A's  money  is  10  %\    125  %\  25  %;    31|  ^; 
431  ^;   62^  ^;   75  %;   100  ^;   150  %;   200  ^;   225 
^  ;  375  %\   1000  %   more  than  B's;   then  B's  is  how 
many  %   less  than   A's? 

Ans.  QT'T;    1U;    20;    23H;    302§;    387r^  ;    42f;    50 
60;  663;  69T33;  78j| 

24.  If  Ahas5$;  1 

33^;  45^;  50$;  683  %\  75%;  84$;  98$;  and 
99s  $  fes  money  than  B;  then  B's  money  is  how  many 
%  more  than  A's  ? 

Ans.  5T\;   1714;   21i73;   293V;    331;  42$;  50;  81T9T: 
100;  220;  300;  525;  4900;  79900  %. 


<208  RAY'8   HIGHER   ARITHMETIC. 


25.  What  %  of  a  number  is  8  %  of  35  %  of  it? 

Ant.  2j 

SOLUTION.- 780XT'0so  =  2ia  =  250  °floo#  =2$#. 

26.  What  ^  of  a  number  is  2j  #  of  2 2  $  of  it? 

ylns.  T'S 

27.  What  ^  of  a  number  is  40  %  of  621  #  of  it? 

-4/i5.  25. 

28.  12  $  of  $75  is  what  $  of  $108?          Ans.  8i 

29.  If  A  of  a  ship  be  sold,  what  %  of  it  is  sold  ? 

SUGGESTION. —  T52  of  100^=4lf  $. 

30.  U.  S.  standard  gold  and  silver  is  9  parts  pure  to 
1  part  alloy:  what  %  of  alloy  is  that?  Ans.  10. 

31.  English  standard  gold  is  22  carats  fine:  how  many 
%  of  pure  metal  in  a  sovereign?    (Art.  205.)    Ans.  91| 

32.  One  pound  English  standard  silver  contains  18  pwt. 
of  alloy :  how  many  %  alloy  is  it?  Ans.  71 

33.  How  many  ^  of  a  township  6  miles  square  does  a 
man  own,  who  has  9000  acres?  Ans.  891*5 

34.  How  many  %   of  his  time  does  a  man   lose,  who 
sleeps  7  hr.  out  of  every  24?  Ans.  29g 

35.  How  many  <f0  of  a  quantity  is  40  %  of  25  %  of  it? 
also,  16  %  of  371  %  of  it?    also,  4^  %  of  120  %  of  it? 
also,  2  %  of  80  %  of  661  %  of  it?    also,  I  %  of  36  %  of 
75$  of  it?    also,  §1%  of  221  %  of  96%  of  it? 

Ans.  10,  6,  5,  ITS,  uVo, 

36.  30  %   of  the  whole  of  an  article  is  how  many  % 
of  i  of  it?  A»s.  45. 

37.  25  %  of  §  of  an  article  is  how  many  %  of  |  of  it? 

Ans.  133 
CASE   III. 

ART.  255.  To  find  a  number  when  a  certain  per  cent 
of  it  is  given. 

RULE. — Divide  the  given  number  by  the  rate  per  cent. ;  the 
quotient  will  be  1  per  cent.,  and  100  times  this  will  be  100  per 
cent.,  or  the  number  required. 

PROOF. — Find  the  given  per  cent,  of  the  answer;  this 
must  be  the  same  as  the  given  number. 

A  man  gave  a  beggar  25  ct.,  which  was  21  %  of  his 
money:  how  much  had  he?' 


PERCENTAGE.  209 


SOLUTION. —  Since  25 ct.  is  2^  $,  1  #  is  -  =10ct.,  and  100  % 
«=  100  times  10  ct.  =  1000  ct.  =  $10.  Ant. 

REMARK.— The  work  may  often  be  simplified.     Thus,  25 ct.  =» 

1  2^  K  A  A 

2i  %,  =~.  =  2§rt,  ™A  of  the  number  required;    hence,  f  x,  or 

luo        •*  ™ 

the  whole  of  it  =  40  times  25  ct.  =  $10.  Or,  since  25  ct.  are  2^  %•, 
100  %  =  40  times  2A  ^  =  40  times  25  ct.  =  $10,  which  amounts  to 
multiplying  the  given  number  by  the  quotient  of  100  divided  by  the 
rate  per  cent. 

1.  6|  is  9  %  of  what  number?      Ans.  75. 

2.  $3.80  is  5  %  of  what  sum?          Ans.  $76. 

3.  IT  is  80  °f0  of  what  number?         Ans.  3*2 

4.  16  is  lj  °/0  of  what  number?  Ans.  1066§ 

5.  2  is  45  %  of  what  number  ?        Ans.  4$ 

6.  $72  is  160  %  of  what?  Ans.  $45. 

7.  31 1  ct.  is  15|  %  of  what?  Ans.  $2. 

8.  421  is  I7so  %  of  what?  Ans.  250. 

9.  $10. 75  is  31  %  of  what?          Ans.  $322.50 

10.  $24  is  1661         %  of  what?  Ans.  $14.40 

11.  162  men  are  4|  °f0  of  how  many?       Ans.  3375. 

12.  $2.68f  is  z         %  of  what?         ^4ns.  $537.50 

13.  $19.20  is -ft       %  of  what?  Ans.  $3200. 

14.  $12675  is  240    %  of  what?       Ans.  $5281.25 

15.  $189.80  is  104  %  of  what?         Ans.  $182.50 

16.  16gal.  1  pt.  is  64  %  of  what?   Ant.  262  gal.  2  qt. 

17.  1  oz.  15?  dr.  is  g  %  of  what?       4n*.  14  Ib.  1  oz. 

18.  *  of  1$  is  15-      %  of  what?  Ans.  8. 

19.  10  mi.  7  fur.  36  rd.  is  75  %  of  what  ? 

Ans.  14  mi.  5  fur.  8rd. 

20.  36  men  of  a  ship's  crew  die,  which  is  42$  %  of  tin 
whole:  what  was  her  crew?  Ans.  84  men. 

21.  A  stock-farmer  sells  144  sheep,  which  is  12f  %  of 
his  flock:  how  many  had  he?  Ans.  1125. 

22.  A  merchant  sells  35$  of  his  stock   for   $6000: 
what  is  it  all  worth  at  that  rate?  Ans.  $17142.86 

23.  I  shot  12  pigeons,  which  was  2f  %  of  the  flock: 
bow  many  escaped?  Ans.  438. 

REVIEW.— 255.    What  is  Case  3?    The  rule?    The  proof?     How  maj 
the  work  bo  simplified  ?     Give  an  example. 
18 


210  RAY'S   HIGHER   ARITHMETIC. 


21.  A,  owing  B,  hands  him  a  $10  bill,  and  says,  thera 
is  6|  $  of  your  money:  what  was-thc  debt?  Ans.  $160. 

25.  82.")' i>  G2A  %  of  A's  money,  and  41*  %  of  B's. 
what  did  each  have?  Aw.  A  $40,  B  $60. 

26.  A  found  $5,  which   was   13:1  %   of   whflt  llc   had 
before:  how  much  had  he  then  ?  Ans.  $42.50 

27.  B  lost  a  $3  bill,  which  was  31 J-  %  of  what  he  had 
loft:  how  much  had  he  at  first?  Ans.  $12.60 

28.  I  drew  48  %  of  my  funds  in  bank,  to  pay  a  note 
of  $150:  how  much  had  1  left?  An?.  $162.50 

29.  A  farmer  gave  his  daughter  at  her  marriage  65  A. 
2  R.  26  P.  of  land,  which  was  3  %  of  his  farm:  how  much 
land  did  he  own?  Am.  2188  A.  3  R. 

30.  A  pays  $13  a  month  for  board,  which  is  20  %  of 
his  salary:  what  is  his  salary?  Ans.  $780  a  year. 

31.  Bought  8000  bu.  of  wheat,  which  was  57?  %   of 
my  whole  stock  :  how  much  had  I  before?  Ans.  6000  bu. 

32.  Paid  40  ct.  for  putting  in  25  bu.  of  coal,  which  was 
llf  %  of  its  cost:  what  did  it  cost  a  bu.?    .Ans.  14  ct. 

33.  81  men  are  5  %  of  60  %  of  what?  Ans.  2700  men. 

34.  45  is  20  %  of  18!  %  of  what?  Ans.  1200. 

35.  If  32  %  of  75  %  of  800  %  of  a  number  is  1539, 
what  is  it?  Ans.  8  01 A 

36.  A,  owning  60  %  of  a  ship,  sells  7  2  %  of  his  share 
for  $2500  :   what  is  the  ship  worth?    Ans.  $55555. 55f 

37.  A  father,  having  a  basket  of  apples,  took  out  88i% 
of  them  for  his  children  ;  of  these,  he  gave  37^  %  to  his 
son,  who  gave  20  %  of  his  share  to  his  sister,  who  thus 
got  2  apples:  how  many  were  in  the  basket?      Ans.  80. 

CASE  IV. 

ART.  256.  A  number  being  given,  which  is  a  given  per 
cent,  more  or  less  than  another,  to  find  that  other. 

RULE. — Represent  the  number  required  by  100  per  cent ;  allow 
for  the  given  rate  per  cent,  of  increase  or  decrease;  the  re- 
tulting  rate  per  cent,  will  represent  the  number  given;  and  from 
this,  100  per  cent,  or  the  number  required,  can  be  found  by 
Case  III. 

PROOF. — Find  the  given  per  cent,  of  the  answer,  and 
add  it  to,  or  subtract  it  from,  the  answer,  as  may  be  neces- 
:  thn  rest) It  must  bo  the  same  as  the  given  number. 


PERCENTAGE. 


I  pay  $377  rent  for  my  house,  which  is  16  %  more 
than  I  paid  last  year  :  what  was  the  rent  then  ? 

SOLUTION.  —  Call  the  rent  100 

last  year   100   per  cent.  ;    the  1  6  *                     e 

rent  this  year  being  16  %  more,  —  —  7«  H*  n  n   \  „  , 

is  116  #;  divide  §337  by  116,  to  '     o  Q  JJ                 inn 

get  1  fc,  and  multiply  the  quo-  con              -^^^ 

tient,  $3.25,  by  100,  to  get  100$,  5°^$325.00 
or  the  rent  last  year,  $325. 

I  am  28  yr.  old,  which  is  14?  per  cent,  less  than  riy 
brother's  age  :  how  old  is  he  ? 

SOLUTION.  —  Call  my  brother's  age  100  f0\  my  age  being  14|  % 
less,  is  85  7  %  ',  28  yr.  divided  by  85|  =  g  jj|j,  is  1  %  ;  and  100  tir'Ci 
this,  or  x§6  =  32|,  is  100  ft,  or  my  brother's  age.' 

1.  136  is  20  %  less  than  what?         Ans.  170. 

2.  $4.80  is  33|    %  more  than  what?  Ans.  $8.60 

3.  i  is  50  %  more  than  what?  Ans.  $ 

4.  .  j\  is  28  %  less  than  what?  Am.  3" 

5.  96  da.  is  100     %  more  than  what?     .4ns.  48  da. 

6.  2576  bu.  is  60  %  less  than  what?  Ans.  6440  bu. 

7.  872ct.  is  87s    %  less  than  what?  Ans.  $7. 

8.  1|  is  500  %  more  than  what?  Ans.  7*5 

9.  3  bu.  2,pk.  7  qt.  is  8  %  more  than  what? 

Ans.  3  bu.  1  pk.  637  qt. 

10.  42  mi.  Ifur.  20  rd.  is  55  %  less  'than  what? 

Ans.  93  mi.  6  fur. 

11.  21b.  9  oz.  4gdr.  is  50  %  less  than  what? 

Am.  5  Ib.  2  oz.  9s  dr. 

12.  773  is  99s  %  less  than  what?  ^«s.  155| 

13.  $920.931  is  337s  %  more  than  what? 

Ans.  $210.50 
1£    $4358.061  is  233  J  %  more  than  what? 

Ans.  $1307.41^ 

15.  In  64i  gal.  of  alcohol,  the  water  is  7s  %   of  the 
spirit  ;  how  many  gal.  of  each  ?    Ans.  60  gal.  sp.,  4j  gal.  w. 

16.  A  coat  cost  $32;    the   trimmings  cost  70%   less, 
and  the  making  50  %  less,  than  the  cloth  :  what  did  each 
cost?    Ans.  Cloth  $17.77|,  trim.  $5.  33s,  mak.  $8.88§ 

SUGGESTION.  —  Cloth  =  100  %\  trimming  =  30  %\  making  =  50  % 
their  sum  180  %  =  $32.     Find  1  %  and  then  the  rest. 


«M2  RAY'S   HIGHER   ARITHMETIC. 

17.  If  a  bu.  of  wheat  makes  39.Ub.  of  flour,  and  the 
cost  of  grinding   is   4  °/0;  how  many  bbl.  of  flour  can  a 
farmer  get  for  80  bu.  of  wheat?  Ant.  15  A  bbl. 

18.  How  many  eagles,  each  containing  9  pwt.  16.2  gr. 
of  pure  gold,  can  I  get  for  455. 6538  oz.  pure  gold  at  the 
mint,  allowing  1A  °/0  for  expense  of  coinage?    Ans.  928. 

19.  2047  is  10  %  of  110  %  less  than  what  number? 

*==!,  AM.  2300. 

20.  4246^  is  6  %  of  50  %  of  466!  %  more  than  what 
number?  Ans.  3725. 

21.  A  drew  out  of  bank  40^  of  50%  of  60%  of 
70  %  of  his  money,  and  had  left  $1557.20;  how  much 
had  he  at  first?  Am.  $1700. 

22.  I  gave  away  42f  %  of  my  money,  and  had  left  $2  ; 
what  had  I  at  first?  Atu.  $3.50 

23.  A  man  dying,  left  33g  %   of  his  property  to  his 
wife,  60  %  °f  ^e  remainder  to  his  son,  75  %  of  the  re- 
mainder to  his  daughter,  and  the  balance,  $500,  to  a  ser- 
vant ;  what  was  the  whole  property,  and  each  share  ? 

AM.  Property,  $7500;  wife  had  $2500;  son  $3000; 
daughter  $1500. 

24.  In  a  company  of  87,  the  children  are  37s  %  of  the 
women,  who  are  443  %  °f  the  men  ;  how  many  of  each? 

Ans.  54  men,  24  women,  9  children. 

25.  In  a  school,  5  %  of  the  pupils  are  always  absent, 
and  the  attendance  is  570 ;    how  many  on  the  roll,  and 
how  many  absent?  Ans.  600;  and  30. 


XV.   APPLICATIONS  OF  PERCENTAGE. 

AIIT.  257.    The  importance  of  thoroughly  understanding 

Percentage,  can  not  be  over  estimated. 

• 

In  marking  goods,  in  calculating  gain  or  loss,  the  value 
of  investments,  interest,  commission,  &c.,  its  applications 
are  various,  important,  and  of  daily  occurrence  ;  and  no 
one  can  be  a  ready  and  complete  accountant  until  he  ia 
familiar  with  its  principles  and  methods. 

Although  the  special  rules  are  the  simplest  and  most 
satisfactory,  especially  in  the  first  two  cases,  the  general 

RE  TIB  w.— 256.  What  is  Case  4?  The  role?  The  proof?  SO!M 
example. 


GAIN   AND   LOSS.  213 


rule  (Art.  252),  is  easy  to  recollect,  and  serves  very  well 
for  the  first  three  cases,  and  even  for  the  fourth  case,  after 
Blight  preparation. 

Particular  care  should  be  exercised  in  ascertaining  the 
quantity  which  is  the  standard  of  comparison  on  which 
the  rate  per  cent,  takes  effect;  and  bear  in  mind  that  this 
quantity  always  represents  100  per  cent. 

REMARKS. — 1.  The  words  per  cent,  have  no  reference  to  money. 
All  they  mean  is,  on,  or  by  the  hundred,  and  a  rate  per  cent,  is  a  num- 
ber of  hundredths  of  the  quantity,  whatever  it  may  be.  It  is  true,  the 
phrase  is  used  more  in  speaking  of  money  than  of  any  other  article, 
but  it  is  perfectly  proper  to  say,  "Ten  per  cent,  of  the  labor;" 
"Twenty  per  cent,  of  the  cloth;"  "Seven  per  cent,  of  the  time;" 
meaning  BO  many  hundredths  of  the  labor,  cloth,  and  time,  re- 
spectively. 

2.  The  advantage  of  using  rates  per  cent.,  or  ratios  in  hundredths, 
instead  of  ordinary  ratios,  is  that  rates  per  cent.,  like  all  other 
fractions  having  a  common  denominator,  are  more  easily  compared 
than  ordinary  ratios.  Thus,  it  is  not  easy  to  see  which  of  the  two 
ratios  7  ^nd  ^  is  the  greater,  but  as  soon  as  they  are  expressed  in 
hundredths,  viz:  42f  %  and  44g  %,  the  difficulty  vanishes. 


XVI.  GAIN  AND  LOSS 

ART.  258.  The  increase  or  decrease  which  any  variable 
quantity  undergoes,  is  called  its  gain  or  loss. 

The  rate  of  gain  or  loss  is  the  rate  per  cent,  the  gain 
or  loss  is  of  the  quantity  on  which  the  gain  or  loss  accrues. 
The  quantity  on  which  gain  or  loss  accrues,  is  the  stan- 
dard of  comparison  in  questions  of  gain  or  loss,  and  is 
therefore  100  per  cent. 

GENERAL  RULE. 

Represent  the  quantity  on  icliich  gain  or  loss  accntes  by  100 
per  cent.,  and  proceed  by  such  rule  of  Percentage  as  the  nature 
tf  the  question  requires. 

NOTE. — In  gain  or  loss  of  money,  the  sum  of  money  invested,  or 
east,  is  the  standard  of  comparison,  and  is  therefore  100  per  cent. 

REVIEW. — 257.  What  single  rule  comprehends  all  the  cases  ?  What 
care  should  bo  exercised  in  all  questions  of  percentage?  What  must  th« 
itandard  of  comparison  always  represent? 


214  RAY'S   HIGHER   ARITHMETIC. 


ART.  259.    There  are  4  cases  of  Gain  or  Loss,  solved 
like  the  4  corresponding  cases  of  Percentage. 

CASE  I. — Given,  the  quantity  on  which  gain  or  loss  accrv^i, 
and  the  rate  of  gain  or  loss,  to  find  the  gain  or  loss. 

Having  invested  $4800,  my  rate  of  gain  is  13s  %' 
what  is  my  gain? 

$4800 

ANALYSIS.— The  $4800  being  13j 

the  quantity  on   which   the   gain  14400 

accrues,  is  100$;  the  gain  being  4800 

13|  ^=13g  hundreths  of   it,   is  4200 

found  as  in  Case  I  of  Percentage,         jj— • 
bj  multiplying  it  by  13|,  and  di-         f  66  6.0  0  Gam. 
Tiding  the  product  by  100. 

$  5  4  6  6  Sum  after  gain. 

REMA.RKS. — When  the  gain  or  loss  is  known,  the  amount  after 
gain  or  loss,  is  obtained  by  an  addition  or  a  subtraction. 

1.  If  my  rate    of  gain  is  25  %,  how  should  I  mark 
goods  for  sale  that  cost  me  $8;    $7.50;    $6.25;   $4.75; 
$3.87A;  $2,62A;  $1.93|;  62^ct.;  15ct.ayard? 

yi»s.$10;  $9.1m;  $7.81i;  $5.93?;  $4. 84§;  $3.28|; 
$2.42T'a;  784  ct.;  18.fct.  a  yard. 

2.  If   I    must    lose  20   %    on    damaged   goods,    how 
should  I  mark  those  that  cost  me  122ct.:  25  ct.;  43|ct.; 
75 ct;  $1.10;  $2. 40;  $3.50;  $4.37i;  $5.81|;  $6.56J; 
$7.68|;  $8.10  a  yard? 

Ans.  10  ct.;  20  ct.;  35  ct. ;  60  ct.;  88  ct.;  $1.92; 
$2.80;  $3.50;  $4.65;  $5.25;  $6.15;  $6.48,  a  yard. 

3.  The  population  of  a  town  was  1760  last  year,  and 
has  increased  26?  %  :  what  is  it  now?  Ans.  2222. 

4.  A  city  containing  42540  inhabitants,  lost  11|  % 
of  them   by  cholera:    how  many  died,  and  how  many  are 
left?  Ans.  4963  died;  and  37577  left. 

5.  A  Ib.  Tr.  contains  5760  gr.,  and  a  Ib.  av.  21 3  e  % 
more:  how  many  gr.  does  it  contain?          Ans.  7000  gr. 

REVIEW. — 257.  What  is  said  of  the  words  per  cent.?  What  advantage 
In  using  rates  per  cent  instead  of  ordinary  ratios?  Give  an  example. 
268.  What  is  gain  or  loss  ?  What  is  the  rate  of  gain  or  loss  ?  In  all 
questions  of  gain  or  loss,  which  quantity  is  100  %1  What  is  the  general 
rule?  In  gain  or  loss  of  money  which  quantity  is  100  J&?  Why? 
259.  How  many  cases  of  gain  or  loss  ?  What  is  Ca*e  1  ?  Analyze  the 
example.  How  can  tho  quantity  nftrr  gain  or  lof>s  be  found  ? 


(JAIN    AND   LOSS.  215 


6.  A  g:>cs  42  mi.  3  fur.  18rd.  a  day;  B,  15  %  faster: 
how  far  does  B  go  per  day?     Am.  48mi.  6fur.  14.7rd. 

7.  TJ.  S.  standard  gold  is  n)  pure,  and  English  stand- 
ard gold  is  l|f  %  purer:  how  pure  is  it?     Ans.  \*  pure. 

8.  U.  S.  standard  silver  is  T9o  pure,  and  English  stand- 
ard silver  is  2  3  °/0  purer:  how  pure  is  it?     Ans.  40  pure. 

9.  The   cost  of  publishing  a   book  is  50  ct.   a  copy; 
if  the   expense   of   sale  be  10  %  of   this,  and  the   profit 
25  c/0'.  what  does  it  sell  for  by  the  copy?      Ans.  67^  ct. 

10.  A  began  business  with  $5000  :  the  1st  year  he 
gained  14|  %,  which  he  added  to  his  capital;  the  2d  year 
he  gained  8  %,  which  he  added  to  his  capital;  the  3d  year 
he  lost  12  %,  and  quit:  how  much  better  off  was  he  than 
when  he  started?  Ans.  $452.92 

\\f  A  bought  a  farm  of  government  land  at  $1.25  an 
acre;  it  cost  him  160  %  to  fence  it,  160  c/0  to  break  it 
up.  80  %  for  seed,  100  %  to  plant  it,  100  %  to  harvest 
it,  112  %  for  threshing,  100  %  for  transportation:  each 
acre  produced  35  bu.  of  wheat,  which  he  sold  at  70  ct.  a 
bushel :  how  much  did  he  gain  on  every  acre  above  all 
expenses  the  first  year?  Ans.  $13.10 

12.  What  must  I  sell  a  horse  for,  that  cost  me  $150, 
to  gain  35  %1.  Ans.  $202.50 

13.  A  gave  $4850  for  his  house,  and  offers  it  for  20  % 
less:  what  is  his  price?  Ans.  $3880. 

14.  Bought  hams  at  8  ct.  a  Ib. ;  the  wastage  is   10  %: 
how  must  I  sell  them  to  gain  30  %1     Am*  llf  ct.  a  Ib. 

SUGGESTION. — Since  a  Ib.  wastes  10  %,  or  J'Q,  I  get  only  -fa  Ib 
for  Set.,  and  a  Ib.  at  that  rate  costs  8§  ct.;  to  which  30  %  must  be 
added,  to  get  the  selling  price. 

15.  I  bought  .a  cask  of  brandy  containing  46  gal.  at 
$2.50   per  gal.;  if  6 gal.  leak  out,  how  must  I    sell   the 
rest,  so  as  to  gain  25  %?  Ans.  $3.59!  per  gal. 

16.  I  started  in  business  with  $10000,  and  gained  20 
%  the  first  year,  and  added  it  to  what  I  had;  the  2d  year 
I  gained  20  %,  and  added  it  to  my  capital;  the  3d  year  I 
gained  20  %.     What  had  I  then?  Ans.  $17280. 

ART.  260.  CASE  II. — Given,  the  quantity  on  which  gain 
or  loss  accrues,  mid  the  gain  or  toss,  to  find  thf  rate  of 
gain  or  loss. 


>1«  RAY'S   IHGIIIOK   ARITHMETIC. 

REMARK.  —  If  the  quantity  on  which  gaiu  or  loss  accrues,  and 
the  quantity  after  gain  or  loss,  are  given,  their  difference  is  the  gain 
or  loss,  and  the  question  would  come  under  this  case. 

The  population  of  Cincinnati  in  1840  was  45000,  and 
in   1850   was   165000  ;    how  many  per  cent,  had    it   in 
creased  in  the  interval? 

ANALYSIS.—  Here  the  gain  is  120000,  which,  compared  wit 
45000,  the  quantity  on  which  the  gain  accrues,  id  V/oVu0  —  \  = 
of  100$  =266|  %.  Ans. 

1.  If  I  double  my  money,  how  many  per  cent,  do  I 
gain?  Ans.  100. 

2.  If  I  lose  half  my  goods,  how  many  per  cent,  do  I 
lose?  Ans.  50. 

3.  If  I  buy  at  $1  and  sell  at  §9,  how  many  per  cent 
do  I  gain  ?  Ans.  800. 

4.  If  I  buy  at  $1  and  sell  at  $4,  how  many  per  cent. 
de  I  gain?  Ans.  300. 

5.'    If  I  buy  at  $4  and  sell  at  $1,  how  many  per  cent. 
do  I  lose?  Ans.  75. 

6.  If  I  sell  |  of  an   article  for  what   the  whole  cost 
me,  how  many  per  cent,  do  I  gain  ?  Ans.  80. 

7.  If  I  sell  g  of  an  article  for  what  f  of  it  cost  me. 
how  many  per  cent,  do  I  lose?  Ans.  31  f 

8.  If  a  person  sell  14  oz.  of  candles  for  a  pound,  how 
many  per  cent,  does  he  gain?  Aits.  14? 

9.  How  many  %  larger  is  the  earth's  equatorial  diam- 
eter (15850  mi.)  than  its  polar  diameter  (15798mi.)? 

Ans.  3  %  nearly. 

10.  If  I  sell  an  article  for  3  of  its  cost,  how  many  per 
3ent.  do  I  lose?  Ans.  66  j 

11.  The  U.  S.  half  dollars  coined  since  1853,  contain 
8pwt.   of   standard   silver;    those   coined   before,  contain 
8pwt.  14:}-  gr.  of  standard  silver;  how  many  per  cent,  more 
valuable  are  the  latter  than  the  former?  Ans.  7,~ii 

12.  A  log  1ft.  6  in.  thick,  is  sawn  into  13  boards  each 
1.1  in.  thick  :  what  %  is  wasted?  Ans.  9|t 

13.  The  U.  S.  wine  gallon  contains  231  cu.  in.,  and  the 
beer  gallon  282  cu.  in.:  how  many  %  larger  is  the  lattei 


R  E  v  i  E  w.—  260.  What  i?  Case  2  ?  The  rule  ?  If  tho  quantity  on  which 
gain  or  loss  accrues,  and  the  quantity  after  gain  or  loss,  arc  given,  how  do 
wo  proceed  ?  Analyze  tho  example. 


GAIN   AND   LOSS.  21? 


than  the  former?  and  how  many  %  smaller  is  the  former 
than  the  latter?      A  us.  227^  %  larger;    184'?  %  smaller. 

14.  The  U.  S.  dry  gallon  or  half  peck,  contains  2G8.8 
cu.  in.:  how  many  c/0   larger  is  it  than  the  wine  gallon? 
and  how  many  %  smaller  than  the  beer  gallon? 

Ans.  16i4r  %  larger;  4|?  %  smaller. 

15.  The    imperial    gallon    of    Great    Britain    contain 
277. 274  cu.  in.:  how  many  $,  is  it  larger  than  our  wine 
and  dry  gallons  ?   and  how  many  %  smaller  than  our  beer 
gallon? 

Ans.  20T?h  %  larger;  8TS»%  %  larger;  iVWo  %  less. 

16.  U.   S.    gold  (T9<j  pure)    is   how  many  %   less   pure 
than  English  gold,  (73  pure)?  Am.  lYr 

17.  Gold  22  carats  is  how  many  per  cent,  purer  than 
gold  18  carats  fine?  and  how  many  °/0  pure  is  each? 

Ans.  22g  %  purer;  first,  91 3  °/0  pure;  2d,  75  %'purc. 

18.  If  I  pay  for  a  Ib.  of  sugar,  and  get  a  Ib.  Troy,  what 
^   do  I  lose,   and  what  %    does  the   grocer  gain  by  the 
cheat?  -4ns.  17?  %  loss;  21:lg  %  gain. 

19.  A,  having  failed,  pays  B  $1750  instead  of  $2500, 
which  he  owed  him  :  what  c/0  docs  B  lose?          Ans.  30. 

20.  Sugar  bought  at  6{'ct.  a  Ib.,  is  sold  at  7r.  ct.  a  Ib. : 
what  is  the  rate  of  gain  ?..  -4ns.  14  %. 

21.  An  article-has  lost  20  %  by  wastage,  and  is  sold  for 
40  %  above  cost:  what  is  the  gain  per  cent.?    Ans.  12. 

SOLUTION. —  20  ft,  of  the  cost  is  wasted;  80  %  of  the  cost  re- 
mains available,  on  which  a  gain  of  40  %  is  realized;  40  %  of  80  $ 
=  32  %,  which,  added  to  80  %,  makes  112  fe;  deduct  the  cost,  100  %\ 
the  net  gain  is  12  %• 

22.  If  my  retail  gain  is  33]  %,  and  I  sell  at  whole- 
sale for  10  %  less  than  at  retail,  what  is  my  gain  %   at 
wholesale^        -  Ans.  20. 

23.  Bought  a  lot  of  glass  ;  lost  15  %  by  breakage:  at 
what  %  above  cost  must  I  sell  the  remainder  to  clear  20  c/0 
on  the  whole?  Ans.  41  y3? 

24.  If  a  bu.  of  corn  is  worth  35  ct.  and  makes  2>gal. 
of  whisky,  which  sells  at  24  ct.  a  gal.,  what  is  the  profit 
of  the  distiller?  Ans.  71 1  %- 

ART.  261.  CASE  III. — Given,  the  gain  or  toss,  and  the 
rate  of  c/'tin  or  loss,  to  find  the  quantity  on  which  gui/i  or 
loss  accrues. 

19 


RAT'S  HIGHER  ARITHMETIC. 


By  selling  a  lot  for  34*  %  more  than  1  gave,  my  gain 
is  $423.50  :  what  did  it  cost  me? 

ANALYUIS.—  Since  34|  %  =$423.50,  1  %  =  $423.50  -j-  JMj,  =s 
$12.32  ;  and   100  %,  or  the  whole  cost  =  100  times  §12.32  =  $1232 
as  in  Case  A  of  Percentage. 

REMARK.  —  After  the  quantity  on  -which  gain  or  loss  accrues  :• 
I  IOWH,  (he  gain  or  loss  may  be  added  to  it  or  subtracted  from  it, 
tt  get  the  quantity  after  gain  or  loss  accrues. 

1.  How  large  sales  must  I  make  in  a  year  at  a  profit 
of  8  %  to  clear  $2000?  AH*.  $25000. 

2.  I  lost  §50  by  selling  sugar  at  22-J  %  he-low  cost: 
what  was  the  cost?  Am.  $222.  22'! 

3.  If  I  sell  tea  at  131  %  gain,  I  make  lOct  a  Ib.r 
how  much  a  pound  did  I  give?  Ans.  75  ct. 

4.  1  lost  a  2  2  dollar  gold  coin,  which  was  7?  %  of  all 
I  had:  how  much  had  I?  Ans.  $35. 

5.  A  and  B  each  lost  $5,  which  was  2?  %  °f  A's  an^ 
3s  %  of  B's  money:  which  had  the  most  money,  and  how 
much?  Ans.  A  had  $30  more  than  B. 

6.  I  gained  this  year  §2400,  which  is  120  %  of  my 
pain    last  year,  and   that   is   44g  %  of  my  gain  the  year 
before:  what  were  my  gains  the  two  previous  years? 

Ant,.  $2000  last  year.;  $4500  year  before. 

7.  The  dogs  killed  40  of  my  sheep,  which  was  4^  % 
of  my  flock:  how  many  had  I  left?  Ans.  920. 

ART.  262.  CASE  IV.  —  Given,  (he  quantify  after  gain  or 
lots  h<is  accrual,  and  the  rate  of  gain  or  loss,  to  jind  the 
quantity  on  which  gain  or  loss  accrues. 

Sold  goods  for  $25.80,  by  which  I  gained  7  2  %'•  what 
was  the  cost? 

BOLDTIOV.  —  The  cost  is  TOO  %\  the  $25.80  being  1\  %  more,  it 
107.J  %\  then  1  %  =  $25.80  -=-1  07  .J  =  24  ct.,  and  100  %  or  the 
tost  =  100  times  24  ct.  =  $24  ;  as  in  Case  4  of  Percentage. 

REMARK.  —  After  the  quantity  on  which  gain  or  loss  accrues  is 
fouL-1,  the  difference  between  it  and  !he  quantity  after  gain  or  loss 
will  be  the  gain  or  loss. 


RBvtKw.—  261.  What  is  Ca.«e  3d?    Analyte  the  exhiiple.     Ho  v  can  the 
quantity  after  lose  or  gain  be  found?     262.  What  is  Case  4th? 
Vhe  example.     How  ia  the  gain  or  loss  found  T 


GAIN   AND   LOSS.  21'J 


1.  Sold  cloth  at  S3. 85   a  yd.;  my  gain    wa?   10%: 
how  much  a  yard  did  I  pay?  Am.  $3.50 

2.  Gold   pens,  sold   at  $5   a  piece,  yield    a   profit   of 
33'  %:  what  did  they  cost  a  piece?  AH*.  $3.75 

3.  Sold  oat  for  $952.82  aud  lost  12  %\  what  was  the 
cost?  and  what  would  I  have  got  if   I  had  sold  out   at  a 
gain  of  12  %?  An*.  $1082.75  and  $1212.68 

4.  Sold  my  horse  at  40  %  gain;    with  the  proceeds  I 
bought    another,  and    sold    him    for    $238,   losing  20  %: 
what  did  each  horse  cost  me? 

An*.  $212.50  for  1st,  $297. 50  for  2d. 

5.  Sold  flour   at  an  advance   of  13s  %;   invested   the 
proceeds    in    flour  again,  and  sold  this    lot  at  a   profit  of 
24  %,  realizing    $3952.50:    how  much  did  each   lot  cost 
me?  Aus.  l*t  lot,  $2812.50;  2d  lot,  $3187.50 

6.  An  invoice  of  goods  purchased  in  New  York,  cost 
me  8  %  for  transportation,  aud  I   sold   them  at  a  gain   of 
16s  %  on  their  total  cost  on  delivery,  realizing  $1200: 
what  were  they  invoiced  at?  An*.  $1000. 

7.  The   population  of  a  village  increased  50  %   each 
year  on  the  previous  one,  for  four  successive  years,  and  at 
the  end  of  the  5th  was  405:    what  was  it  at    the  end  of 
each  previous  year?  Am.  80,  120,  180,  270. 

8.  For  6  years  my  property  increased  each  year  on 
the  previous,  100  %,  and  became  worth  $100000:   what 
was  it  worth  at  first?  Ans.  $1502.50 

9.  A  lost   at  play  50  %  of  his  money  the   1st  game, 
50  %  of   the  remainder   the   2d,  and  so   on  for   10  suc- 
cessive  games,  when   ho  was  reduced   to  his  last  dollar: 
what  Lad  he  at  first?  Ans.  $1024. 


XVII.  COMMISSION  AND  BROKERAGE. 

ART.  263.    Ono  who  buys  or  sells  property,  makea  in 
vestments,  collects  debts,  or  transacts   other  business   foi 
the  benefit,  and  at  the  advice  of  another,  is  a  commi'&aion 
merchant,  agent,  or  factor. 

When  the  commission  merchant  lives  in  a  different 
country  or  part  of  the  country  from  his  employer,  he  ia 
frequently  called  correspondent  or  consignee;  goods  pen! 


R  E  v  i  •  w. — 263.  What  ia  a  commigiinn  merchant,  agent,  or  factor  r 


2-»0  RAY'S  HIGHER  ARITHMETIC. 


to  him  to  be  sold  arc  called  a  consignment,  and  the  person 
Bonding  them  a  conniijnor. 

The  charge  made  hy  a  commission  merchant  for  trans- 
acting another's  business,  is  called  his  commi^inn,  and  is 
estimated  at  a  certain  rate  per  cent,  of  the  sum  invested 
or  realized  for  the  other's  benefit.  This  rate  per  cent,  is 
the  rate  of  commission. 

BROKERAGE  is  a  charge  of  the  same  nature  as  commis- 
sion, but  generally  smaller. 

ART.  264.  The  amount  of  the  sole,  purchase,  or  collection, 
is  tl>e  standard  of  comparison,  and  is,  therefore,  100  %. 

The  net  proceeds  of  a  sale  or  collection  is  the  sum  left 
after  deducting  the  commission  and  other  charges. 

GENERAL  RULE  FOR   COMMISSION. 

Represent  by  100  per  cent,  the  quantity  on  which  the  commis- 
sion or  brokerage  is  charged,  {which  is  always  the  amount  of  the 
sale,  purchase,  or  collection,')  and  then  proceed,  by  such  rule  of 
Percentage  as  the  nature  of  the  question  requires. 

Commission  has  4  cases,  solved  like  the  corresponding 
cases  of  Percentage. 

CASE  I. 

ART-.  285.  Given,  the  amount  of  the  sale,  purchase,  or 
collection,  and  the  rate  of  commission,  to  find  the  com- 
mission. 

A  commission  merchant  makes  sales  during  a  year 
amounting  to  $1684'75.37i,  on  which  his  charge  wag 
2-i  %:  how  much  did  his  commission  come  to? 

ANALYSIS.  —  The  amount  of  sales,  $108475.  37A,  is  100$,;  the 
commission,  being  1\  fc,  is  found  by  multiplying  by  2A,  and  dividing 
by  100,  as  in  Case  I  of  Percentage;  this  gives  $4211.88. 

REMARK.  —  If  the  commission  ($4211.88)  be  subtracted  from  the 
amount  of  sales  ($10847o.37A),  the  remainder  ($104203.40^)  will  be 
the  amount  paid  to  the  consignors. 

1.     I  collect  for  A,  $268.40,  and  have  5  %  commis- 
ion:  Low  much  do  I  pay  over?  Ans.  $254.98 


2G3.  What  is  a  consigner?  A  consignment?  A  consignee? 
Flow  is  n.  commission  merchant  paid  f  What  is  the  rate  of  commission  ? 
What  i«  brokerage? 


COMMISSION    AND   BROKERAGE. 


2.  I   sell   for  B,  650  bbl.  of  flour,  at   $7.50  a  bbl, 
28bbl.  of  whisky,  35  gal.  each,  at  22A  ct.  a  gal.:  what  is 
my  commission  at  2\  %?  Ans.  $114.65 

3.  Received  on  commission  25  hhd.  sugar  (36547  lb.): 
of  which  I  sold  10  hhd.  (16875  lb.)  at  6  ct.  a  lb.,  and 
6  hhd.  (8246  lb.)  at  5  ct.  a  lb.,  and  tho  rest  at  54  ct.  a  lb. 
what  is  my  commission  at  3  %?  Ans.  $61.60 

4.  A   lawyer   charged   8  %   for  collecting  a   note  of 
$648.75:  what  is  his  fee,  ana  the  net  proceeds? 

Ana.  $51.90,  and  $596.85 

5.  A  lawyer,  having  a  debt  of  $1346.50  to  collect, 
compromises    by  taking    80  %,  and  charges  5  %   for   his 
fee  :  what  is  his  fee,  and  the  net  proceeds? 

AM.  $53.86,  and  $1023.34 

G.  Bought  for  C,  a  carriage  for  $950,  a  pair  of  horses 
for  $575,  and  harness  for  $120;  paid  charges  for  keeping, 
packing,  shipping,  &c.,  $18.25;  freight,  $36.50:  what 
was  my  commission,  at  3s  %,  and  the  whcle  bill? 

Ans.  $54.83,  and  $1754.58 

NOTE.  —  Commission  is  charged  only  on  the  amount  of  the  purchase. 

7.  Sold  500000  lb.  of  pork,  at  5j  ct.  a  lb.  :  what  is 
my  commission,  at  12;  %1  Ans.  $3437.50 

8.  An  insurance  agent  has  10  %  of  all  sums  received 
for  his  company:  what  docs  he  make  in  a  year,  if  he  re- 
ceives for  the  company,  $28302.75  ?     Am.  $2830.27-1 

9.  An  insurance  agent  has  5  %  of  all  sums  received 
for  his  company,  and  5  c/0  of  what  remains  at  the  end  of 
the  year  after  payment  of  losses:  what  will  he  make,  if  he 
receives  for  his  company,  $47363.87-,  and  pays   losses. 
$31344.50?  Ana.  $3169.16 

10.  An  architect  charges  3j  c/c  for  designing  and  super- 
intending a  building,  'vhich  cost  $27814.60:  what  is  his 
fee?  Ans.  $973.51 

11.  A  factor  has  2?  %  commission,  and  3i  c/0  for  guar- 
anteeing payment:  if  the  sales  are  $6231.25,  what  doe» 
he  get  r  Am.  $389.45 

RK  VIEW.  —  264.  What  quantity  is  tho  standard  of  comparison  ?  What 
docs  it  represent  then  ?  What  are  the  net  proceeds  of  a  sale  or  collection  ? 
What  is  the  general  rule  for  conm.ission  and  brokerage?  How  muny  cases 
of  commission  ?  265.  What  is  Caso  1  ?  Analyze  tho  example.  What  ara 
the  net  proceeds  ? 


;>i>'2  RAY'S    IIIGHKR    ARITHMETIC. 


]•_'.  An  architect  charges  \\  %  for  plans  and  specifics 
lions,  and  2,3  %  for  .superintending:  what  dues  lie  make, 
if  the  building  costs  $14902.50 '(  Ans.  $014.73 

i:>.      A  broker  sells  for  me  lOhhd.  sugar  (9256  lb.),  at 
5  ct.  a  lb.;  what  is  his  brokerage,  at  f  r/n,  and  my  proceeds? 
Ans.  Brokerage,  $3.47;    proceeds,  $459.33 

14.  A  sells  a  house  and  lot  for  me  at  $3850,  and  charges 
t~%  brokerage:  what  is  his  fee?  Ans.  $24.00.1 

15.  I  have  a  lot  of  tobacco  on  commission,  and  sell  it 
through  a  broker  for  $4(342.85:  my  commission  is  2-1  %, 
the   brokerage  \\  c/0:  what  do  I  pay  the  broker,  and  what 
do  I  keep?         Ans.  I  keep  $63.84;  and  brok.  $52.23 

K).  What  does  a  tax  gatherer  get  for  collecting  a  tax  of 
$-37850,  at  3  %,  and  how  much  does  he  pay  over? 

Ans.  $1135.50  rec'd;  $30714.50  paid  over. 
17.     A  tax  collector  is  paid  4i  %  for  collecting  a  tax  of 
$218090.75:  what  is  his  fee,  and  the  net  proceeds? 

AM.  $9814.35  fee,  $208282.40  paid  over. 

CASE  II. 

ART.  268.  Given,  the  commission,  and  the  amount  of 
the  sale,  purchase,  or  collection,  to  iind  the  rate  of  com- 
misnion.  ( 

1.  An  auctioneer's  commission  for  selling  a   let  was 
$50,  and  the  sum  paid  the  owner  was  $1200:  what  was 
the  rate  of  commission?  Ans.  4  c/0, 

SUGGESTION. — Find  the  amount  of  the  sale,  $12oO;  then  find 
what  </c  $50  is  of  $1250,  as  in  Case  II  of  Percentage. 

2.  A  commission   merchant  sells  800   bbl.  of  flour,  at 
$6.43?  a  bbl.,  and  remits  the   net   proceeds,    $5021.25: 
what  is  his  rate  of  commission?  Ans.  2j  %. 

3.  The  cost  of  a  building  was  $19017.92,  including 
the  architect's  commission,  which  was  $553.92:  what  rato 
did  the  architect  charge?  AUK.  3  %. 

4.  Bought  flour  for  A;  my  whole  bill  was  $5802.57 
including   charges,    $70.85,   and    commission,    $148.72 
find  the   rate  of  commission.  Ans.  2ij  %. 

5.  Charged   $52.50   for  collecting  a  debt  of  $1050: 
what  was  my.  rate  of  commission?  ^l//s.  5  '/t  . 

0.     An    agent    gets   $169.20   for  selling   property  fur 
$8400:  what  was  his  rate  of  brokerage?  Aits.  2  %. 


COMMISSION    AND   BROKERAGE. 

7.  My  commission  for  selling  books  was  §6.92,  and 
the  not  proceeds,  $02.28:  what  rate  did  I  charge? 

An*.  10  %. 

8.  Paid  $38.40  for  selling  goods  worth  $0400 :  what 
wis  the  rate  of  brokerage?  .  Ans.  jj  f/0. 

9.  Paid   a   broker   $24.16,  and  retained  as  my  part 
:>f  (he  commission   $42.28,  for   selling  u  consignment  at 
$2116:  what  was  the  rate  of  brokerage,  and   my  rate  of 
commission  ?  Ans.  Brok.  1  %;  Com.  *2i  </#* 

111.  A  tax  gatherer  is  paid  $3711  for  collecting  a  lax 
of  $74220:  what  rate  is  allowed  him?  Ans.  5  %. 

11.  A  tax  collector  retains  $0826.45,  and  pays  over 
$539280.55:  what  c/0  is  his  commission?  Ann.  \\%. 

CASE  ITI. 

AUT.  267.  Given,  the  commission,  and  rate  of  com.,  to 
find  the  sum  on  which  commission  is  charged. 

NOTE. — After  finding  the  sum  on  which  commission  is  charged, 
subtract  the  commission,  to  find  the  net  proceeds,  or  add  it,  to  find 
the  whole  coat,  as  the  ease  may  be. 

1.  My  commissions  iu  1  year,  at  2\  %,are  $3500:  what 
wore  the  sales,  and  the  whole  net  proceeds? 

AUK.  $140000  and  $136500. 

SUQOKSTION.—  2^  =  $3-300;  find  1  %,  then  100  %\  as  in  Case 
III  of  Perec ut age. 

2.  An   insurance    agent's   income   is    $1733.45,   being 
10  %  on   the  sums  received  fur  the  company:  what  were 
the  company's  net  receipts?  Ans.  $15001.05 

3.  A    pork    merchant  charged   15  c/n    commission,    and 
cleared  $2370.15,  after  paying  out  SI  200. 75  for  all  ex- 
penses of  packing:  how  many  pounds  of  pork  did  he  pack, 
if  it  cost  4«  ct.  a  pound?  Ans.  530800  Ib. 

4.  An  aj:cnt   purchased,  according  to  order,  10400  bu. 
jf  wheat ;  liis  commission,  at  1]  %,  was  $150,  and  chargos 
for  storage,  shipping,  and  freight,  $527  10*.  what  did  he 
|  ay  a  bushel?  and  what  was  the  whole  cost? 

Am.  $1.20  a  bu.,  and  $13103.10,  whole  cost. 

5.  Pai-1  $04.05  for  soiling  enffoe,  which  was  I  %  brolc* 
eiagc  .  what  are  the  net  proceeds?  Auk.  $7255.95 


REVIEW.— 266.    What  is  Case  2?    2t>7.  What  if  Caa«  5?    How  ar« 
the  not  prococdd  Ibund?     lluw,  the  whole  cost? 


HAY'S   HIGHER   ARITHMETIC. 


0.  lU'coivcd  produce  on  commission  at  2j  r/f,\  my  surplus 
commission,  after  paying  i  ('/0  brokerage,  is  $107.03:  what 
was  the  amount  of   the  sale,  the   brokerage,  and  net  pro 
cceds?yl«s.  Sale,  $01 10;  brok.,  $30.58;  pro.,  $5078.39 

7.  A  is  paid   $801.27  for  collecting  taxes  at  2 A  %'• 
what  were  the  taxes,  and  the  sum  received  by  the  State? 

Ans.  Tax,  $34450.80,  paid  to  State,  $33589.53 

8.  Paid  A    $1952.04   for  collecting,    at   1  .?-<£:   what 
were  my  net  proceeds?  Ans.  $109020.7.9 

CASE  IV. 

ART.  268.  Given,  the  rate  of  commission,  and  the  net 
proceeds,  or  the  whole  cost,  to  find  the  sum  on  which  com- 
mission is  charged. 

NOTES. — 1.  After  the  sum  on  which  commission  is  charged  is 
known,  find  the  commission  by  subtraction. 

2.  When  the  divisor  in  this  case  is  little  less  than  100,  use  the 
contracted  method  of  Division.  (Art.  69.) 

1.  A  lawyer  collects  a  debt  for  a  client,  takes  4  ^  for 
his  fee,  and   remits  the  balance,  $207.00:  what  was  the 
debt  and  the  fee?  Ans.  $210.25  and  $8.05 

SOLUTION. — The  debt  being  the  quantity  on  •wjich  commission 
is  charged,  is  100  %\  if  4  %  is  taken,  there  is  left  90  %  =  §207.00: 
find  1  f0  and  then  100  fa]  as  in  Case  IV  of  Percentage. 

2.  Sent  $1000  to  buy  a  carriage,  commission  2j  %  : 
what  must  the  carriage  cost?  Ans.  $975.01 

SUGGESTION.— 100  %  4~2i  %  =  102|  %  =$1000;  find  1  % 
then  100  %. 

3.  A  buys  per  order  a  lot  of  coffee;  charges,  $50.85; 
commission,  l.{  %;  the  whole  cost  is  $539.01:  what  did 
the  coffee  cost?  Ans.  $470.80 

SUGGESTION. — Take  out  the  charges;  the  rest  as  before. 

4.  Buy  sugar  at  2]  $  commission,  and  2  j  %  for  guar 
ntccing  payment:  if  the  whole  cost  is  $1500.  what  wa 

the  cost  of  the  sugar?  Ans.  $1431.98 

SUGGESTION.— 100-J-2]  -f  2i  =  104f  ^=$1500;  rest  as  before 

KEVIRW.— £6S.  AVhat  is  Case  4  ?  How  can  the  commission  be  found 
If  tJO  divisor  is  little  less  than  100,  what  may  bo  done? 


STOCKS   AMD   DIVIDENDS. 


5.  Sold  2000  hams  (20672  lb.);  commission,  1\  %, 
guarantee,  2?  %,  net  proceeds  due  consignor,  $2448. o4: 
what  did  the  hams  sell  for  a  lb.?  Ans.  12A  ct. 

G.  What  tax  must  be  assessed,  to  yield  $20782.45 
net  proceeds,  and  pay  the  collector  3g  %',  and  what  is  the 
collector  paid?  AM.  Tax,  $27646.40;  col.,  $863.95 

7.  What  tax  must  be  raised,  to  yield  $1044073.50 
nd  pay  for  collection,  at  §  %1  Ans.  §1050640. 

8.  A  sells  1000  bbl.  (30468|  gal.)  of  whisky;  broker 
age,  i  %;   proceeds,  $7254:   how  much   a  gallon  was  it 
sold  for  ?  -4ns.  24  ct. 

9.  Sold  cotton   on   commission   at  5  %\   invested  the 
net  proceeds  in  sugar,  commission,  2  %;  my  whole  com- 
mission was  $210 :  what  was  the  value  of  the  cotton  and 
sugar?  Ans.  Cotton,  $3060  ;  sugar,  $2850. 

SUGGESTION. — Cotton  =  100^.  Com.  on  Cotton  =  5$.  Pro- 
ceeds =  95  % .  Com.  on  sugar  =  T^  cf  05  %  (Art.  256)  —  \\\%. 
Whole  Com.  =  5^  +  l|f  %  =C|f  %  =  $210;  find  1  %,  KC. 

10.  Sold  flour  at  3-j  %  commission;  invested  §  of  its 
value  in  coffee,  at  1A  %  commission;  remitted  the  balance. 
$432.50:  what  was  the  value  of  the  flour,  the  coffee,  and 
my  commissions?         Ans.  Flour,  $1500;  Coffee,  $1000; 
1st  Com.,  $52.50;  2d  Com.,  $15. 

11.  Sold  a  consignment  of  pork,  and  invested  the  pro- 
ceeds in  brandy,  after  deducting  my  commissions,  4  %  for 
selling,  and  \\%  for  buying.    The  brandy  cost  $2304.00  : 
what  did  the  pork   sell   for,   and  what  were  my  commis- 
sions ? 

Ans.  Pork,  $2430;  1st  Com.,  $97.20;  2  Com.,  $28.80 

12.  Sold  1400  bbl.  of  flour,  at  $6.20  a  bbl.;  invested 
the  proceeds  in  sugar,  as  per  order,  reserving  my  commis- 
sions, 4  %  for  selling,  and  lA  %  for  buying,  and  the  ex- 
pense of  shipping,  $34.16:    how  much   did    I  invest  in 
sugar?  Ans.  $8176 


XVIII.   STOCKS  AND  DIVIDENDS. 

Anr.  269.  A  joint-stock  company  is  an  association  of 
individuals,  empowered  to  transact  a  specified  business 
under  certain  restrictions. 


2-20  RAY'S   HIGHER   ARITHMETIC. 


The  business  transacted  by  them  is  generally  such  as  to  exceed 
Hie  menus  of  one  person;  as,  Banking,  .Mining,  Insurance,  &c. 
Railroads,  canals,  steamboats,  turnpikes,  bridges,  telegraphs,  £c., 
are  owned  and  managed  by  joint-stock  companies. 

The  capital  of  a  company  is  called  its  stock,  and,  for 
coTivenie.iice  is  usually  divided  into  shares  of  §100,  or  §50, 
for  each  share  a  certificate  is  issued. 

Persons  who  own  shares  are  called  stockholders,  or  sharc- 
Jioli/cm ;  stock  can  be  transferred  from  one  person  to 
another,  the  certificates  being  evidence  of  ownership. 

The  dividend  is  the  gain  to  be  divided  among  the  stock 
holders,  in  proportion  to  their  amounts  of  stock.     Hence, 
a  joint-stock  company  is  in  the  nature  of  a  partnership. 

On  account  of  the  great  number  of  shares  in  such  a 
concern,  it  is  convenient  to  declare  the  dividend,  as  a 
certain  rate  per  cent,  of  the  whole  stock  ;  this  rate  per 
cent,  may  be  called  the  rate  of  dividend. 

CASE  I. — Given,  the  stock,  and  rate  of  dividend,  to  find 
the  dividend  for  that  slock.  (See  Case  I,  Percentage.) 

1.  I  own    18  shares  of   $50  each,  in    the  City  Insur- 
ance Co.,  which  has  declared  a  dividend  of  7;j  %  :   what 
do  I  receive?  Ans.  §07.50 

2.  I  own  147  shares  of  railroad  stock  (§50  each),  on 
which    I    am    entitled    to  a    dividend  of  5  %,  payable  in 
stock:  how  many  additional  shares  do  I  receive? 

Ans.  7  shares,  and  §17.50  toward  another  share. 

3.  A  has  218  shares  bank  stock  (§100  each),  and  gets 
a  dividend  of  12  %:  how  much  is  that?       Ans.  §'201  <j. 

4.  The  Cincinnati  Gas  Co.  declare  a  dividend  of  18  %: 
what  do  E  get  on  50  shares  (§100  each)?     An*.  §1*00. 

5.  The  Western  Stage  Co.  declare  a  dividend  of  4i  per 
cent      if  their  whole  stock  is  §150000,  how  much  is  dis- 
tributed to  the  stockholders?  Aim.  $(?750. 

it.  A  railroad  Co.,  whoso  stock  account  is  §4-~)l>OOf|, 
declared  a  dividend  of  3^  c/0'.  what  sum  was  distribut.  I 
,  monir  the  stockholders?  Ans.  §148'Jl>0. 


II  K  v  I  K  w. — 2'if.  What  U  a  j'-int-stock  company  ?  What  kind  «'f  luisi. 
Tii'?."  d»  iln-y  transact?  What  ii  the  .<tc,ck  ?  lluw  L<  it  dividi-d  ?  What  arc 
the  rtofkiioldi-rs?  What  is  the  dividend?  How  is  it  divided  among  llu' 
itock holders?  What  is  the  rate  of  dividend?  What  is  Case  I? 


PAR,    DISCOUNT,    AND   PREMIUM. 


7.  A  Telegraph  Co.,  with  a^  capital  of  $75000,  declares 
a  dividend  of  7  %,  and  ha?  $0500  surplus:    \vl»;it   has  it 
ea  rued  ?  vi  us.  §  1  1750. 

8.  I  own  24  shares  of  stock  ($25  each)  in  a  Fuel  Co. 
which   declares  a  dividend  of  0  %;   I  take  my  dividend   in 
coal,  at  Set.  a  bu.:  how  much  do  I  get?          Aut.  450  hu. 

A  IIT.  270.  CASE  It.  —  Given,  (he  stock,  and  dividend,  to 
find  (he  rale  of  dividend.  (See  Case  II,  Percentage.) 

1  My  dividend  on  72  shares  bank  stock,  (§50  each), 
is  $324:  what  was  the  rate  of  dividend?  Aim.  9  c/0. 

2.  A    Turnpike  Co.,  whose    stock    is    $225000,  earns 
$10384.50  :   what  rate  of  dividend  can  it  declare? 

Ans.  7  %,  and  $0-34.50  surplus. 

3.  The  receipts  of  a  Canal  Co.,  whose  stock  is  $3050000, 
in  one  year  arc  $250484;  the  outlay  is  $70383:  what  rate 
of  dividend  can  it  declare?    Ans.  4i  %,and  $12851  sur. 

ART.  271.  CASE  III.  —  The  dividend,  and  rate  of  did- 
di'int,  given,  to  Jind  (he  stuck  corresponding.  (See  Case  III, 
Percentage.) 

1.  An  Insurance  Co.  earns  $1  8000,  and  declares  a  15% 
dividend:  what  is  its  stock  account?          Ans.  $120000. 

2.  A  man  gets  $04.50  as  a  7  %  dividend:   how  many 
shares  of  stock  ($50  each)  has  he?  Ans.  27. 

3.  Received  5  shares  ($50  cach\  and  $26  of  another 
blhi  re,  as  an    8  c/0   dividend  on    stock  :    how  many  shares 
had  I  ?  An*.  Gi>. 

AUT.  272.    CASE  IV.—  (See  Case  IV,  Percentage.) 

1.  Received,  10  %   stock  dividend,  and   then   had  102 
shares  ($;>0  each),  and  $15  of  another  share:   how  many 
shares  had  1  before  the  dividend?  Ans.  03. 

2.  Having  received  two  dividends  in  stock,  one  of  5%, 
another  of   8  %,  my  stuck    has  increased  to  507  shares: 
hu»v  many  had  I  at  lirst  ?  Ans.  500. 


XIX.   TAR,  DISCOUNT,  AND  PREMIUM. 

AHT.  273.  PAH,  DISCOUNT,  and  PKEMITM,  are  mercan- 
tile terms,  applied  tit  nunn-i/^  s/oc'/.-s,  lnjinl*,  and  «/M/*/*. 

Mitnry  is  the  circulating  medium  of  trade,  in  the  lorn:  of 
gold  and  silver  coins,  and  bank  notes. 


228  HAY'S    HIGHER   ARITHMETIC. 

S/orks,  are  money  invested  in  Banks,  Insurance  compa- 
nies, itc.,  in  the  form  of  shares-  $50  or  $100  each. 

Drafts,  Li'lls  of  exchange,  or  checks,  are  written  orders  for 
money.  Jionds  arc  written  obligations  to  pay  money  at  a 
future  time. 

ART.  274.    All  money.  sfoc7is,  bonds,  and    drafts,  have 
value  on  their  face,  called  nominal  or  par  value. 

Their  real  value,  is  what  they  arc  intrinsically  worth 
and  sell  for. 

When  their  real  is  the  same  as  their  nominal  value 
they  arc  said  to  be  par,  (the  word  par  meaning  equal,  in 
Latin).  When  they  sell  for  less  than  their  nominal  value, 
they  arc  below  par,  or  at  a' discount.  When  they  sell  for 
more  than  their  nominal  value,  they  are  above  par,  or  at 
a  premium. 

ART.  275.  Discount  is  how  much  less,  money,  stocks, 
drafts,  &c.,  arc  worth,  than  their  face.  Premium  is  how 
much  more,  they  are  worth,  than  their  fa ce. 

Rate  of  premium,  or  rate  of  discount,  is  the  rate  per  cent, 
the  premium  or  discount  is  of  the  face. 

ART.  276.  The  face  or  par  value  of  money,  stocks, 
drafts,  £c.,  is  the  standard  of  comparison :  hence,  this 

GEXERAL   RULE. 

Represent  the  face  by  100  per  cent ;  and  then  proceed  by  such 
rule  of  Percentage  as  the  nature  of  the  question  requires. 

This  subject  has  4  cases,  solved  like  the  4  corresponding 
cases  of  Percentage. 

CASE  I. —  Given,  the  par  value,  and  tlie  rate  of  premium 
or  discount,  to  find  the  premium  or  discount.  (Sec  Case  I 
Percentage.) 

NOTE. — If  the  result  is  a  premium,  it,  must  be  added  to  the  par 
'o  get  (he  real  value;  if  it  is  a  discount,  it  must  be  subtracted. 


RE  VIE  w.— 270.  What  is  Case  2?  271.  Case  3?  272.  What  doo 
Cnso  4  correspond  to?  273.  What  arc  Par,  Discoim*,  and  Premium 
What  is  money  ?  What  nro  stocks  ?  What  are  drafta  7  Whut  <>th-T  name 
do  they  have?  What  are  bonds?  274.  What  is  the  par  value  of  mo.i-v 
Ktoclc:,  Ac.?  What  is  their  real  value?  When  nro  they  par?  Wl-v  so 
called?  When  arc  they  below  par,  or  at  a  discount?  When  nro  tlu>y 
above  par,  or  at  a  premium  ?  275.  What  is  discount  ?  Premium?  TUta 
»f  discount?  Rato  of  premium? 


PAR,    DISCOUNT,    AND   PREMIUM.  22tf 


1.  Buy  18  shares  stock  ($100)  at  8  %  discount:  find 
the  discount  and  cost.  A  us.  $144,  and  $1656. 

2.  Sell  the  same  at  4]  %  prcm.:  tind  the  prein.,  the 
price,  and  gain.  Ana.  §81,  $1881,  and  $225. 

3.  Bought  G2  shares   Railroad  stock  ($50)  at  28  % 
premium:  what  did  they  cost?  AM.  $3908. 

4.  What  is  the  cost  of  47  shares  Railroad  stock  ($50) 
at  30  %  discount?  Am.  $1645. 

5.  Bought  $150  in  gold  at  \  c/0  premium:  what  is  the 
premium,  and  cost?  Ans.  $1.12i  and  $151. 12V 

6.  Sold  a  draft  on  New  York  of  $2568.45,  at  A  % 
premium:  what  do  I  get  for  it?  Ans.  $2581.29 

7.  Sold  $425  uncurrcnt  money  at  3  %  discount:  what 
did  I  get,  and  lose?  Ans.  $412.25  and  $12.75 

8.  What  is  a  $5  bank  note  worth  at  6  c/0  discount? 

Am.  $4.70 

9.  Exchanged  32  shares  Bank  stock  ($50),  &%  pre- 
mium, for  40  shares  Railroad  stock  ($50),  10  %  discount, 
and  paid  the  difference  in  cash:  what  was  it?  Ans.  $120. 

10.  Bought  98  shares  stock  ($50)  at  15  %  discount; 
cave  in  payment  a  hill   of  exchange  on  New  Orleans  for 
$4000    at  £  %  premium,  and   the  balance   in   cash:  how 
much  cash  did  I  pay?  Ans.  $140. 

11.  Bought  56  shares  Turnpike  stock  ($50)  at  69  %; 
sold  them  at  76A  %  :  what  did  I  gain?  Ans.  $2JO. 

REMARK. — The  cost  of  stocks  is  generally  given  as  so  many  <f0 
of  the  face,  instead  of  so  many  %  discount  or  premium. 

12.  Bought  telegraph  stock  at  106  %\  sold  it  at  01  %\ 
what  was  my  loss  on  84  shares  ($50)?  Ans.  $630. 

13.  What  is  the  difference  between  a  draft  on  Philadel- 
phia of  $8651.40,  at  \\%  premium,  and  one  on  N.Orleans 
for  the  same  amount,  at  A-  %  discount?      Ans.  $151.40 

14.  Bought  18  shares  Railroad   stock  ($50)  at  12  % 
discount,  paying  A  c/0  brokerage:  what  did  it  cost  me? 

Ans.  $795.96 

15.  Find  the  cost  of  9  Ohio  State  bonds  (§500)  at  8  % 
premium,  \  %  brokerage.  Ans.  $4896.45 

REVIEW. — 27i>.  What  is  the  standard  of  comparison?  What  is  the 
general  rule  for  buying  or  selling  money,  stocks,  drafts,  Ac.?  How  many 
cases  in  this  subject?  What  do  they  correspond  to?  What  is  Caso  H 
Hoiw  is  the  real  value  found  after  the  premium  or  discount  is  known  '( 


HAY'S   HIGHER   ARITHMETIC. 


10.  T  change  $380  hank  notes  for  gold  ;it  1  .1  ^  pro- 
iniuin:  $25  of  the  notes  arc  1  '/£,  discount;  $40  are  2  '/^ 
difeount;  !?(.>•>  are  a'^  discount;  and  $10  are  '&f/fl  discount: 
il'  I  receive  S-170  in  gold,  how  much  change  should  be 
given  the  broker?  Ann.  1  5  ct. 

17.  Buy  3o4  shares  .stock  ($50),  at  16  f/f/  discount 
^brokerage  ;'  %\  sell  them  at  10  %  premium,  broke  rn^a 
f"  %:  what  \*  my  gain?  Ans.  $4524.52 

AKT.  277.  CASK  IT.  —  Gicen,  f  he  face,  and  the  discount 
or  premium,  t<>  find  the  rale  of  discount  or  premium.  (See 
Case  II,  Percentage.) 

NOTES.  —  1.  If  the  fitce  ami  the  real  value  are  known,  take  their 
difference  for  the  discount  or  premium;  then,  apply  the  rule. 

2.  If  the  rate  of  g<iin  or  lost  is  required,  the  real  value  or  cost  is 
the  standard  of  comparison,  not  the  face. 

1.  Paid  $2401.30  for  a  draft  of  $2300  on  New  York: 
what  was  the  rate  of  premium?  Ans.  1?  %. 

2.  Paid  §2508.03  for  26   shares  stock   ($100),  and 
brokerage,  $25.03:   what  the  rate  of  discount? 

•Aiu.  4}  %. 

3.  Bought  112  shares  Railroad  stock  ($50)  for  $3040  : 
what  was  the  rate  of  discount?  AUK.  35  '^  . 

4.  If  the  stock   in  last  example  yields  8  c/0  dividend, 
what  is  my  rate  of  gain  f  Aim.  12  fV  %. 

5.  I  sell  the  same  stock  for  $5936:  what  rate  of  pro 
mium  is  that?  what  rate  of  gain?       Ans.  6^;   63  \*%. 

G.  If  I  count  my  dividend  as  part  of  the  gain,  what 
is  my  rate  of  gain  ?  ^l;/.s.  75  f^  %. 

1.  Exchanged  12  Ohio  bonds  ($1000),  1%  premium, 
for  280  shares  of  Railroad  stock  ($50):  what  rate  of  dis- 
count were  the  latter?  Ans.  87  $,. 

8.  Gave  $266.661  of  notes,  4  %  discount,  for  $250 
of  gold:  what  rate  of  premium  was  the  gold?  Ans.  2|  %. 

!).  Bought  58  shares  Mining  stock  ($50)  at  40  %  prc- 
fenvnn,  and  gave  in  payment  a  draft  on  Boston  for  $4000: 
vh;it  rate  of  premium  was  the  draft?  Ans.  lj  %. 

10.    Received  $4.60  for  an  uncurrent  $5   note:    what 
was  the  rate  of  discount?  Ans.  8  '. 


EV/.  —  277.  What  is  Case  2?  What  case  of  Percent;iyo  iloc?  il 
corruspond  to?  What,  if  the  face  and  the  real  value  nro  known  f  What. 
If  the  rule  of  gain  or  loss  i*  roquirc'l  ? 


PAR,   DISCOUNT,  AND   PREMIUM. 


ART.  278.  CASE  TIT.  —  Given,  (he  (Hwm/nt  or  premium, 
ami  flu1  rate  of  discount  orjfcTMMMN}  to  find  tlte  face.  (See 
Case  III,  Percentage.) 

NOTKS.  —  1.  After  the  face  is  obtained,  add  to  it  the  premium,  or 
subtract  the  discount,  to  get  the  real  value  or  cost. 

'2.   It'  (lie  gain  or  los.t  is  given,  arid  the  rate  per  cent,  of  the  foci 
orrespoudiug  to  it,  work  by  Case  111,  Percentage. 

1.  Paid  36  ct.  premium  for  gold  |  %  above  par:  how 
much  gold  was  there?  Ans.  $48. 

'1.  Took  stock  at  par;  sold  it  for  il\  %  discount,  and 
lost  $11  7:  how  many  shares  ($50)  had  1?  Ans.  104. 

3.  The  discount,  at  7  2  %,  on  stocks  was  $93.75:  how 
many  shares  ($50)  were  sold?  Ans.  25. 

4.  Buy   stock   at  4-i  %  premium  ;   sell   at   8.J  %  pre- 
mium; gain,  $345:  how  many  shares  ($100)?  Ans.  92. 

5.  Buy  stocks  at  14  %  discount;  sell  at  3]  %  pretn.; 
gain,  $11)2.50;  how  many  shares  ($50)?  Ans.  22. 

G.  The  premium  on  a  draft,  at  g  ^,  was  $10.30:  what 
was  the  face?  Ans.  $1184. 

7.  Buy  stocks  at  6  %  discount;  sell  at  42  c/0  discount; 
loss,  $600:  how  many  shares  ($50)?  AUK.  37. 

8.  Stocks   at   12  c/o  discount,  brokerage,  $0.03,  cost 
$239.97  less  than  the  face:  how  many  shares  ($50)? 

Ans.  41. 

9.  Bonds,   at   20  %   premium,    brokerage    g  %,    cost 
$300.87'  more  than  the  fa-.-e:  what  is  the  face? 

Ans.  $1450. 

10.  Buy  uncurrent  bank  notes  at  10  %  discount,  2*  % 
brokerage;   sell  them  at  par,  and  gain  $348.75:  what  was 
the  face  of  the  notes?  Ans.  $4500. 

11.  Buy  stocks   at   40  %   discount,   brokerage    \\  %\ 
sell  them  at  20  %   discount,   brokerage    1.}  %,  and    gain 
$574.  87A:  how  many  shares  ($50)?  Aits.  63. 

AllT.  279.  CASE  IV.  —  Given,  the  real  value,  and  (he  rate 
of  premium  or  tlis'-ount,  to  find  the  face  of  the  drafts,  stock, 
&c.  (Sec  Case  IV,  Percentage.) 

NOTES.  —  1.  After  the  face  ia  known,  take  the  difference  between 
it  and  the  real  value,  to  find  the  discount  or  premium. 


RKVIKW. — 278.  What  is  Cnso  3  ?  What  docs  it  correspond  to?  How 
can  tho  real  value  or  cost  bo  found?  When  tho  gain  or  Ion  it  given,  aoJ 
the  rate  per  cont.  of  the/ricr,  how  proceed? 


KAY'S  HIGHER  ARITHMETIC. 


2.  Bear  in  mind  that  tho  rale  of  premium  or  discount,  and  the 
rate  of  gain  or  loss,  are  entirely  different  things;  the  former  is  re- 
ferred to  the  par  value  or  /ace,  as  a  standard  of  comparison,  the 
latter  to  the  real  value  or  cost. 

1.  What  is  the   face  of  a  draft  on   Baltimore   costing 
2861.45,  at  li  %  premium?  AM.  $2819.16 

2.  Invested  $1591  in  stocks   at   26  %  discount:    how 
lany  shares  ($50)  did  I  buy?  Ans.  43. 

8.  Bought  a  draft  on  New  Orleans  at  3  c/0  discount,  for 
$0398.80:  what  was  its  face?  Ans.  $6430.45 

4.  Notes  at  65  %  discount,  2  %  brokerage,  cost  $881  .  79  : 
what  is  their  face  ?  Ans.  $2470. 

5.  Exchanged  17  Railroad  bonds  ($500),  25  %  below 
par,  for  bank  stock  at  6]  c/0  premium:  how  many  shares 
($100),  did  I  get?  Ans.  60. 

G.  How  much  gold,  at  g  %  premium,  will  pay  a  check 
for  $7567?  Ans.  $7520. 

7.  How  much  silver,  at  1}  %  premium,  can  be  bought 
for  $3252.96  of  currency?  Ans.  $3212.80 

8.  How  largo  a  draft,  at  j  %  premium,  is  worth  54  city 
bonds  ($100),  at  12  %  discount?  Ans.  $4740.15 

9.  Exchanged  72  Ohio  State  bond's  ($1000),  at   Gj  % 
premium,  for   Indiana    bonds  ($500),  at    2  %   premium: 
how  many  of  the  latter  did  I  get?  Ans.  150. 


XX.   INSURANCE. 

ART.  280.  INSURANCE  is  of  two  kinds  —  Insurance  on 
property  and  Life  Insurance;  the  latter  will  be  explained 
Under  the  head  of  Annuities. 

Insurance  on  Property  is  of  two  kinds — Fire  and  Marine; 
he  former  takes  effect  upon  fixed  property,  as  houses  and 
heir  contents;  the  latter  applies  to  property  transported 
ty  water,  as  vessels  and  their  cargoes. 

R  E  v  i  E  w. — 279.  What  is  case  4  ?  What  docs  it  correspond  to?  How 
is  the  discount  or  premium  found?  What  is  the  distinction  between  tho 
rate  of  premium  or  discount,  and  the  rate  of  gain  or  loss?  What  is  the 
standard  of  comparison  for  the  former?  What  for  the  latter?  2SO.  How 
aiany  kinds  of  insurance?  How  is  insurance  on  property  divided?  What 
U  Fire  insurance  ?  Marine  ? 


INSURANCE. 


Property  conveyed  by  railroads  is  insured  after  the 
manner  ol'  marine  insurance. 

ART.  281.  Insurance  on  property,  then,  is  security 
against  loss  by  fire  or  the  dangers  of'  transportation. 

The  companies  or  individuals  tb.at  guarantee  against 
BXi^h  loss,  arc  called  underwriters,  or  insurers.  If  an  indi- 
vidual insures,  it  is  called  out-door  insurance. 

The  written  contract  between  the  insurers  and  the  in- 
sured is  called  the  policy. 

The  sum  charged  for  insuring  is  called  the  premium, 
and  is  a  certain  rate  per  cent,  of  the  amount  insured. 
This  rate  per  cent,  is  called  the  rate  of  insurance. 

Insuring  property  is  called  taking  a  risk. 

ART.  282.  Insurance  has  4  cases  solved  like  the  4  cor- 
responding cases  of  Percentage.  The  amount  insured  is 
the  standard  of  comparison;  hence,  this 

GENERAL  RULE   FOR    INSURANCE. 

Represent  ilie  amount  insured  by  100  per  cent.,  and  then  pro- 
ceed by  such  rule  of  Percentage  as  the  nature  of  the  question 
requires. 

CASK  I.  —  Given,  the  rate  of  insurance,  and  the  amount  in- 
sured, to  find  the  premium.  (Sec  Case  I,  Percentage.) 

1.  Insured  a  house  for  $'2500,  and  furniture  for  §000, 
at/a^,:  what  is  the  premium?  Ans.  $18.60 

2.  Insured  g  of  a  vessel  worth  $24000,  and  3  of  itg 
cargo  worth   $36000,  the  former  at  2|  %,  tta   latter  at 
li<  $:  what  is  the  premium?  Ans.  $607.50 

3.  What  is  the  premium   on  a  cargo  of    railroad    iron 
worth  $28000,  at  1?  %1  Ans.  $490. 

4.  Insured  goods  invoiced  at  $32*760,  for  3  mon.,  at  j8o 
%:  what  is  the  premium?  Ans.  $262.08 

5.  My  house  is  permanently  insured  for  $1800,  by  a  de- 
posit of  10  annual  premiums,  the  rate  per  year  being  f  %: 

REVIEW.—  280.  What  is  said  of  property  conveyed  by  railroad?,  Ac.? 
281.  What  is  insurance  on  property?  Who  are  underwriters?  What  is 
out-door  insurance?  What  is  the  policy?  Tho  premium?  The  rate  of 
insurance?  What  is  insuring  property  called?  282.  How  many  cases  in 
Insurance?  What  id  the  standard  ol  comparison?  What  ia  tho  general 
mlo  ?  What  is  Case  1  ?  What  d<>o»  it  correspond  to  ? 
20 


RAY'S   HIGHER   ARITHMETIC. 


how  much  did  I  deposit?  and  if,  on  terminating  the  insur- 
ance, 1  receive  my  deposit  less  5  %,  how  much   do  I  get  ? 
A/is.  $135  deposited  ;   §128.25  received. 

G.  A  shipment  of  pork  costing  $1275,  is  insured  at 
g^.thc  policy  costing  75  ct.:  what  docs  the  insurance 
cost?  .  J//s.  £7. hi •} 

7.  An  Insurance  company  having  a  risk  of  $25000  at 
rao#>,  re-insured  $10000  at  4%  with  another  office,  and 
$5000  at  1$  with  another:  how  much  premium  did  if 
clear  above  what  it  paid?  Ans.  $95. 

ART.  283.  CASE  IT. —  Given,  tlie  amount  insured,  ana 
premium,  to  find  the  rate  of  insurance.  (Sec  Case  II,  Per- 
centage.) 

1.  Paid  819.20  for  insuring  |  of  a  house  worth  $4800 : 
what  was  the  rate?  Ans.   i  f/0. 

2.  Paid  $234,  including  cost  of  policy,  $1.50,  for  in- 
suring a  cargo  worth  $18000  :  what  was  the  rate? 

AM.H  %. 

3.  Bought  books  in  England  for  $2408  ;   insured  them 
for  the  voyage  for  $46.92,  including  the  cost  of  the  policy, 
$2.50  :  what  was  the  rate?  Ans.  \\%.' 

4.  A  vessel  is  insured  for  $42000;  $18000  at  21  %, 
•Si  5000  at  3£  %.  and  the  rest  at  4j  f/0  :  what  is  tho  rate 
on  the  whole  $42000?  An*.  -V\  %. 

5.  I   took   a   risk   of  $45000;    re-insured   at  the  same 
rate,    $10000  each,  in   3  offices,  and   §5000    in   another; 
my  share    of  the    premium  was~$202.50:    what    w;»s   the 
rate  ?  (   AUK.  2.;'  %. 

0.  I   took  a  risk  at  \\.%\   re-insured  \  of  it  at   2  %, 
and  4  of  it  at  2i  %  :  what  rate  of  insurance  do  I  get  on 
what  is  left?  Ans.  -ft  %. 

SUGGESTION. —  Find  what  per  cent,  of  the  risk  is  paid  for  re- 
insurance; take  this  from  my  rate,  and  find  what  per  cent.  »Uo  re- 
mainder is  of  the  partial  risk  taken  by  me. 

ART.   284.     CASE  III. —  Given,  the  premium,  and  rate  of 
insiu-anee,  to  find  the  amount  insured.     (Sec  Case  III.  I'or 
ccntage.) 

1.  Paid    $118   for   insuring,   at    5  %  :    what    was    TJC 
amount  insured?  J//s.  $14750. 


RHVIKW. — 233.  What  is  C;iso  2  ?      AVhat  docs  Case  2  corrospon  ' 
ZSJ-.    Whtit  is  Case  3?     What  doe*  it  corro.sjximl  to? 


INSURANCE. 


2.  Paid  $411. 37i  for  insuring  goods,  at  1  \  %  :  what 
was  their  value?  .!//.•>•.  $'21-\'2~t. 

X.  1'aid  $42.30  ftir  insuring  I  of  my  house,  at  ,'o  '/£  : 
what  is  the  house  worth?  .  I  //.s.  $7:">-0. 

4.  Took  a  risk   at  2}  $;  re-insured   |  of  it  at  il\  'ff  \ 
my  share  of  the  premium   was   $107.13:    how  large,  was 

he  ri^k?  AHA.  $26284. 

5.  Took   a   risk   at  Is  %\  re-insured  half  of  it  at  tin 
same -rate,  and  \  of  it  at  \\c/f,;  my  share  of  the  premium 
was  $58.11:  how  large  was  the  risk?         AHS  $11(370. 

0.  Took  a  risk  at  2  <fc\  re-insured  $10000  of  it  at 
2'  ^5  and  $8000  at  \\  <:/0  ;  my  share  of  the  premium  was 
$-07. oO:  what  suiu  was  insured?  AHS.  $J8000. 

AUT.  285.  CASE  IV. —  (Hrm,  flic  mlr  nf  imwrancr;  find 
tin'  ttinnttnt  of  property  to  bn  Insured,  to  jind  flic  tini'-nnt  to 
be  titxi/i'i'J  so  us  to  coccr  b'tth  property  and  pri-mium. 

In  this  case,  the  amount  insured  is  m<tiJv  up  nf  the 
property  and  premium;  so  that  if  a  loss  occurs,  both  the 
value  of  the  property  insured,  and  the  premium,  shall  be 
recovered. 

The  value  of  the  property  is  a  certain  rate  per  cent.  7-ss 
than  the  amount  insured,  since  it  is  less  than  that  amount 
by  the  premium,  which  is  always  a  certain  rate  per  cent, 
of  the  amount  insured;  hence,  the.  amount  insured  is  foun  1 
as  in  Case  IV  of  Percentage. 

What  sum  must  he  insured,  to  cover  property  worth 
$4800,  and  premium  at  the  rate  of  3  '/t  ? 

ANALYSIS. — The  ammini  iiisuro'l  ueiug  the  t»tan«lur«l  of  com- 
parison, is  KlO  ft;  tin-  premium  is  .'  %  ;  tho  property.  (!?-lf»lMi),  is  (he 
remaining  W.l  ft:  then  1  ft.  =  S-JSOO  ~-  O'.ii  =  $18.241 '_'-{-,  au<* 
100  ft  is  100  times  this,  =  $J$lM.l'J,  ihe  union ut  to  be  insured. 

It  generally  happens  in  this  case,  that     $4800. 
the   divisor   wants   but   little   of  being  a  £4. 

unit,  and  the  division  can    he  performed  .12 

with  advantage  by  Case  III  of  Contracted     c  j.  o  ?  j~T  9 
Division  of  Decimals  (Art.  1  59).     Ilcncc, 
this  mode  of  (iteration,  when    applied  to   this  case   of  In- 
surance, is  generally  stated  in  the  following  Rule. 

fl  B  v  i  K  w.— 2Sa.  \\"h;it  U  fiiso  4  ?  \Vhnt.  d»i^  it  ccrrc.«pon.I  to?  \\by  7 
Analyze  tho  ex:itnple.  Whut  i-  f:iid  nf  tho  divisor  ?  ilovr  uiuy  the  Jivisiua 
be  perfuruiixl  ?  Apply  it  tu  the  example. 


230  RAY'S   HIGHER   ARITHMETIC. 


PRACTICAL  RULE, 

FOR    INSURING    BOTH    PROPERTY    AND    PREMIUM. 

Find  the  2}rcmnim  on  the  property  to  be  insured;  then,  the 
premium  on  this  premium :  then,  the  premium  on  the  2d  pr& 
miiun,  if  necessary,  and  so  on,  neglecting  all  fyures  below  mills; 
ike  sum  of  the  successive  premiums  will  be  the  whole  premium  ;  to 
Jind  the  amount  to  be  insured,  add  the  property  to  the  premium. 

1.  What  sum  must  be  insured  to  cover  property  worth 
$2600,  and  prcm.  at  TO  %t  Am.  $2618.33 

2.  Insured    to    cover   a    shipment    of    pork,    valued    at 
$12368.50,  at  1  c/0  :  find  the  amt.  insured,  and  prem. 

Ans.  $12493.43,  am'tins'd;  $124.93,  prcm. 

3.  Insured  to  cover  my  library,  $1856.20,   at  Tfi  %  '• 
what  was  the  premium?  Ans.  $11.20 

4.  Insured  to  cover  property  to  the  value  of  $4840,  at 
3  %  :  what  was  the  premium?  Ans.  $36.57 


XXI.    TAXES. 

ART.  286.  TAXES  are  money  paid  by  the  subjects  of  a 
government,  to  defray  its  expenses  ;  and  are  either  direct 
or  indirect. 

Direct  taxes  consist  of  a  property -tax,  and  poll-tax  or 
capilation-lax,  and  are  generally  collected  once  a  year; 
a  prujHThj-tdx  is  levied  on  all  property,  but  that  exempt 
by  law;  it  is  estimated  as  a  certain  rate  per  cent,  of  the 
assessed  value  of  the  property :  this  rate  per  cent,  is  called 
the  rate  of  taxation. 

\  i>nll-ta.r.  or  <-<tpitation-tax  is  a  fixed  sum,  charged  on 
every  citizen,  without  regard  to  his  property. 

This  subject  has  four  cases,  solved  like  the  four  corre- 
sponding cases  of  Percentage:  the  taxable  property  is  the 
standard  of  comparison.  Hence,  this 

GENERAL   RULE. 

Represent  the  taxable  property  by  100  ft;  then  proceed  by  suck 
rule  of  1'eivenlage  as  the  nature  of  the  question  requires. 


UEVIKW.— Wh.it  is  t!:o  practical  rule  ?  2SC.  What  are  taxes  ?  What 
tre  direct  taxes  7  What  is  a  property-tax?  How  is  it  estimated?  What 
is  a  poll-tax  ?  What  is  tho  general  rulo? 


TAXES. 


CASE  I.  —  Given,  the  taxable  property,  and  the  rale  nf  tax- 

ation, to  find  the  property  -tax.      (See  Case  I,  Percentage.) 

fc 
NOTE.  —  If  there  is  a  poll-tax,  the  sum  produced  by  it  should  be 

added  to  the  property-tax,  to  give  the  whole  tax. 

1.  The  taxable  property  of  a  county  is  $486250,  and 
the  rate  of  taxation  is  78  ct.  on  $100;    that  is,  j68o  %: 
what  is  the  tax  to  be  raised?  Am.  $3792.75 

11  KM.  —  The  rate  of  taxation  being  usually  small,  is  expressed 
most  conveniently  as  so  many  cents  on  $100,  or  as  so  many  mills 
on  $1. 

2.  A's  property  is  assessed  at  $3800  ;  the  rate  of  taxa- 
tion is  96  ct.  on  $100  (iVo  %)  :  what  is  his  whole  tax,  if 
he  pays  a  poll-tax  of  $1  ?  Ans.  $37.48 

3.  The   taxable  property  of  Cincinnati,   in   1855,  was 
$85330880,  on  which  was  charged  a  tax  of  TSO*O  %  for  the 
State,  74oVo  %  f°r  the  county,  and  ToVo  %   f°r  tne  town 

a  %  j 


much  was  raised  for 


ship  and  city:  in  all, 
each  and  for  all  ? 

Ans.  State  tax,  $273058.82;  county,  $353269.84; 
township  and  city,  §636568.36;  total,  $1262897.02 

In  making  out  bills  for  taxes,  a  table  is  used,  containing 
the  units,  tens,  hundreds,  thousands,  &c.,  of  property, 
with  the  corresponding  tax  opposite  each. 

To  find  the  tax  on  any  sum  by  the  table,  take  out  the 
tax  on  each  figure  of  the  sum,  and  add  the  results.  In 
this  table,  the  rate  is  Ij  %,  or  125  ct.  on  $100. 

TAX  TABLE. 


Prop.|  Tax. 

Prop.  |  Tax. 

Prop.  |  Tax. 

Prop.  1  Tax. 

Prop.  |  Tax. 

$1 

.0125 

$10 

.125 

$100 

$1.25 

$1000 

$12.50 

$10000 

$125 

2 

.025 

20 

.25 

200 

2.50 

2000 

25. 

20000 

250 

8 

.0375 

30 

.376 

300 

3.75 

3000 

37.50 

30000 

375 

4 

.05 

40 

.50 

400 

5. 

4000 

60. 

40000 

500 

6 

.OG25 

50 

.025 

500 

G.25 

5000 

G2.50 

50000 

G25 

6 

.075 

GO 

.76 

600 

7.50 

GOOO 

75. 

00000 

750 

7 

.0875 

70 

.875 

700 

8.75 

7000 

87.50 

70000 

875 

Q 

.10 

80 

1. 

800 

10. 

8000 

100. 

80000 

1000 

9 

.1125 

90 

1.125 

900 

11.25 

9000 

112.50 

90000  \l-2:> 

4.    AVhat  will  be  the  tax  by  the  table,  on  property  as- 
sessed at $25349? 


K  i:  v  i  K  w. — What  is  Case  1  ?     To  what  does  it  correspond  ? 


•jbS  RAY'S   HIGHER   ARITHMETIC. 

So  LunoN.  —  The  tax  for  20000  is  250;  for  &000  is  02.50;  for  300 
is  3. 75;  for  40  5s  .60;  for  9  is  .ll'JP,  which,  added,  give  the  tax  foi 
$^034!)  -  --  S3JG.3G. 

5.    Find  tho  tax  for  $6815.30.  Ans.  $85.19 

NOTE. — The  tax  for  cents  is  not  in  the  table,  as  being  too  insig- 
nificant; if  desired,  it  may  be  found  from  the  dollars,  by  moving 
be  point  to  the  left,  2  figures:  the  tax  for  30 ct.  =  $.00375. 

0.  What  is  the  tax  on  property  assessed  at  $10424.50 
and  two  polls,  at  $1.50  each?  Ans.  A 


ART.  287.  CASE  II. —  Given,  the  taxable  property,  ami  the 
lax,  to  fiini  the  rate  of  taxation.  (See  Case  II,  Percentage.) 

NOTE. — If  there  is  a  poll-tax,  find  what  it  produces,  and  subtract 
it  from  the  whole  tax  ;  the  remainder  is  the  property-tax,  and  is 
used  to  find  the  rate  of  taxation. 

1.  Property,  assessed  at  $2604,  pays  $19.53  tax  :  wha* 
is  the  rate  of  taxation  ?        Ans.  \  %   =75  ct.  on  $100. 

2.  The  taxable  property   in   a    town   of  1742   polls,    is 
$6814320.    A  tax  of  §66913.54  is  proposed.     If  a  poll- 
fax  of  Sl.25  is  levied,  what  should  be  the  rate  of  taxation? 
,'Sce  Note.)  A -us.  -iQQf/o  =  95  ct.  on  §100. 

3.  Au  estate  of  $350000  pays  a  tax  of  $5670:   what 
's  the  rate  of  taxation?     Ans.  \'it>  *%  =  $1-62  on  $100. 

4.  A't-  tax  is  $53.46;  hv  pays   for  3  polls,  at  $1.50 
?ach,  ami  owns  $8704  taxable  property  :  what  is  tho  rate 
of  taxation?  Ans.  &  %  =  56|  ct.  on  $100. 

ART.  2S8.     CASE  III. —  Given,  the  (ax,  and  the  rate  of 

fa.rtitinii.   to  find  the  assessed  value   of  (he  property.      (See 
Case  III,  Percentage.) 

NOTE. — If  any  part  of  the  tax  arises  from  polls,  it  should  be  first 
deducted  fioin  the  given  tax. 

1.  What  is  the  assessed  value  of  property  taxed  $66.96 
it.  lH  %1  An*.  $3720. 

2.  A  corporation  pays  $564.42  tax,  at  the  rate  of  rVo  %t 
or  46  ct.  on  $100:  find  its  capital.  Ans.  $122700. 

H  K  v  i  K  v:  .--If  trie-re  i*  n  poll-tax,  what  should  be  d-me  ?  2^7.  What  is 
Case  2?  Wlmt  does  it  eorivspond  to?  If  any  part  of  the  tax  is  produced 
from  polls,  what  should  be  done? 


DUTIES  OR  CUSTOMS. 


.i.  A  is  taxed  §7l.Gl  more  than  B;  the  rate  is  Ijs^, 
a.  $1.32  on  §100:  how  much  is  A  assessed  tuorc  than  15? 

An*.  §5425. 

4.  A  tax  of  §4000  is  raised  in  a  town  containing  1024 
polls,  by   a   poll-tax   of  §1,  and   a   propcrfcy-tax  of  iVfl  r/0 
(24  wi.  on  §100):   what  is  the  value  of  the  taxable  pio- 
perty  in  it?  An*.  §1240000. 

5.  A's  income  is  16  %  of  his  capital ;  he  is  taxed  '2\  <fa 
of  his  income,  and  pays  §20.04  :  what  is  his  capital? 

Ans.  §0510. 

ART.  239.     CASE  IV.— (Sec  Case  IV  of  Percentage.) 

1.  A   pays  a  tax  of  1  ./a  #  (§1.35   on   §100)   on    his 
capital,  and  1*1*1  left  §125127. Ht>  :   what  was  his  capital, 
and  hi.s  tax  't          Ait*.  Cap.,  §120840  ;  tax,  §1712.34 

2.  Sold  a  lot  fjr  §7599,  which  was  cost  and  2  %  beside 
paid  for  tax:  vriijt  was  the  cost?  Aits.  $>7450. 


XXII,  DUTIES  OR  CUSTOMS. 

ART.  290.  I-!;/TIES  or  CUSTOMS,  arc  taxes  upon  foreign 
merchandise,  paij  by  the  importer;  they  arc  of  two  kinds, 
ad  valorem  and  .tpccijic. 

Ail  valorem  duties  arc  so  much  on  the  value  of  the 
goods  as  shown  by  the  invoice. 

A<1  valorem  is  Latin  for  on  the  value.  An  invoice  is  a  bill  of  the 
goods,  showing  the  kinds,  quantities,  nnd  cost. 

Specific  duties  arc  a  Gxod  sum  for  a  fixed  quantity, 
without  regard  to  cost;  as,  §10  a  cwt.,  §30  a  hhd.,  &c. 

ART.  291.  In  specific  duties,  certain  allowances  are 
U?ual,  called  draff,  lure,  leakage,  and  breakage. 

Draft  is  an  allowance  made,  that  the  goods  may  hold 
out  when  retailed.  It  is  calculated  as  follows: 


KKVIKW. — 2SS.  What  is  Ca.«e  S?  What  doe*  it  corn-spnml  to ?  Tfnny 
pan  of  tho  t;i.\  if  produced  from  polls,  what  should  he  done!  2Stf  What 
•loog  C;i.so  4  correspond  to?  290.  What  are  Duties  or  Customs?  Uuv 
many  kinds?  What  arc  ad  valorem- duties?  Why  so  called? 


240  RAY'S   HIGHER   ARITHMETIC. 

On  each  parcel  weighing 

1121K,  or  less,  it  is  1  Ib.  from  33Glb.  to  11201b.,  it  is  41b. 

from  11-2  Ib.  to  224  Ib.    ..   21b.  from  1120  Ib.  to  201(3  Ib.,  ..    71b. 

from  224 Ib.  to  330 Ib.   .. .  31b.  over  201  Gib.,  ..    Olb. 

Tare  is  an  allowance  for  the  weight  of  the  hox,  bale, 
cask,  or  whatever  contains  the  goods.  It  is  generally  es- 
timated at  a  certain  per  cent,  on  the  quantity  remaining 
after  the  draft  has  been  deducted. 

Tare  is  sometimes  estimated  at  a  certain  quantity  for  each  cask 
package,  etc.,  or  a  certain  weight  for  each  cwt.,  tun,  &c. 

The  weight  of  the  goods,  before  draft  and  tare  have 
been  allowed,  is  called  gross  weight;  after  they  have  been 
allowed,  it  is  called  net  weight. 

Leakage  is  an  allowance  of  2  $,  on  all  liquors  in  casks 
paying  duty  by  the  gallon. 

Breakage,  usually  10  %,  is  allowed  on  ale,  beer,  and  porter 
en  bottles;  on  other  liquors  in  bottles,  it  is  only  5  %. 

The  common  sized  bottles  are  estimated  at  2|  gal.  per  dozen. 

RULE   FOR   CALCULATING   SPECIFIC   DUTIES. 

ART.  292.  Find  the  net  weight  of  the  goods  in  that  deno 
***iatio>i  for  which  the  rate  of  duty  is  git-en,  and  multiply  it 
ty  the  gicen  rate. 

NOTE. — Observe  that  in  custom-house  business,  28 Ib.  make  1  qr.. 
112  Ib.  make  1  cwt.,  and  2240  Ib.  make  1  tun. 

ART.  293.  In  ad  valorem  duties,  the  calculation  is  one 
of  Percentage,  and  since  the  invoiced  value  of  the  goods 
is  the  quantity  on  which  the  rate  of  duty  takes  effect,  it  is 
100  per  cent. 

RULE   FOR    ALL  QUESTIONS  IN   DUTIES   AD  VALOREM. 

Represent  the  invoiced  value  of  the  goods  by  100  per  cent., 
and  proceed  by  such  rule  of  Percentage  as  the  case  requires. 

REVIEW.  —  290.  What  is  an  invoice?  What  aro  specific  duties? 
291.  What  allowances  are  made  in  specific  duties?  Wh.it  is  draft?  How 
ia  it  estimated?  Wha't  ia  tare?  How  is  it  generally  estimated?  What 
is  it  calculated  on?  How  is  taro  sometimes  estimated?  What  is  gross 
weight?  Net  weight?  What  is  leakage?  Breakage?  How  is  the  quan- 
tity of  liquor  in  bottles  estimated?  292.  What  is  the  rule  for  calculating 
ipcciGc  duties?  In  eattom-honm  business,  hc;v  many  Ib.  in  a  qr.  ? 


DUTIES  AND   CUSTOMS. 


REMARK.  —  According  to  the  last  tariff-bill  of  the  U.  S.,  passed 
1846,  non  3  but  ad  valorem  duties  are  allowed. 

CASE  I.  —  See  Case  I  of  Percentage. 

WHAT   IS   THE   DUTY 

1.  On  45  casks  of  wine,  of  36  gal.  each,  invoiced  • 
$1  .25  per  gal.,  at  40  %  ad  valorem?       Am.  $793.80 

Allow  for  leakage. 

2.  On  12  boxes  fancy  soaps,  each  98A  lb.,  tare  10  %, 
at  5ct.  a  lb.?  Ans.  $52.65 

3.  On  36  boxes  sugar,  each  weighing  6cwt.  2qr.  181b., 
tare  16  lb.  per  cwt.,  at  2£  ct.  a  lb.?  Ans.  $572.40 

SUGGESTION.  —  Tare  =  IGlb.  per  cwt.  =  j1-^  =  \  of  the  lb. 

4.  On  460  E.  E.  of  broadcloth,  at  $3.20  per  ell,  at 
30  %  ad  valorem?  and  if  I  pay  $41.40  freight,  how  much 
per  yard  should  I  charge  to  gain  20  %? 

Ans.  Duty,  $441.60;  price  per  yd.,  $4.08 

5.  On    50   drums  figs,  each  weighing  57  lb.,  draft  as 
usual,  tare  20  lb.  to  the  cwt.,  at  $12  a  cwt.?  Ans.  $246.43 

6.  .On  8hhd.  of  sugar,  each  12  cwt.  3qr.  141b.,  tar« 
12  lb.  to  the  cwt.,  at  l£ct.  a  lb.?  Ans.  $153.  7ft 

7.  On  an  invoice  of  carpets,  worth  $1859.60,  at  30  <fa 
ad  valorem?  Ans.  $557.88 

8.  On  680  boxes  of  raisins,  invoiced  at  $1.25  a  bo*., 
at  40  %  ad  valorem  ?  Ans.  $340. 

9.  On  26  chests  of  tea,  each  118  lb.,  tare  10  lb.  per 
chest,  at  10  ct.  a  lb.?  Ans.  $275.60 

10.  On  42hhd.  of  sugar,  each   13  cwt.  3qr.,  and  the 
tare  71b.  per  cwt.,  at  $2  a  cwt.?  Ans.  $1077.89 

11.  On  30  bags  of  rice,  each  7  cwt.  2  qr.  10  lb.,  tare 
12  %,  at  $1.20  a  cwt.?  Ans.  $239.30 

12.  On  oil-cloth,  40  yd.  long,  and  3/d.  2  ft.  8  in.  wido 
worth  75  ct.  a  sq.  yd.,  at  30  %  ad  valor*  m?     Ans.  $35. 

ART.  294.    CASE  II.  —  See  Case  //,  Percentage. 

1.    If   goods   invoiced    at    $3684.50   pay    a    duty   of 
$1473.80,  what  is  the  rate  of  duty?  Ans.  40  %. 

REVIEW.  —  293.    \Vhat  ia  the  general  rule  for  questions  in  ad  valorem 
duties?    What  kind  of  duties  only  are  allowed  in  this  country?     What 
does  Case  1  correspond  to  ? 
21 


242  RAY'S   HIGHER  ARITHMETIC. 

2.  If  laces,  invoiced  at  $7018.75,  cost,  when  landed, 
$9142.50,  what  is  the  rate  of  duty?  Ans.  20  %. 

3.  If  40   lihd.  (63  gal.  each)  of  molasses,  invoiced  at 
12$  ct.  a  gal.,  pay  $92.61  duty,  after  allowing  2%  for 
leakage,  what  is  the  rate  of  duty?  •  Ans.  30  %. 

ART.  295-    CASE  III. — See  Case  III,  Percentage. 

1.  Paid  $806.12  duty  on  watches,  at  35  %'.  what  were 
hey  invoiced  at,  and  what  did  they  cost  in  store? 

Ana.  Invoiced,  $2303.20;  cost,  $3109.32 

2.  The  duty  on  1800yd.  of  silk  was  $337.50,  at  25  % 
ad  valorem:  what  was  the  invoice  price  per  yd.?  and  whal 
must  I  charge  per  yd.  to  clear  20  %? 

,4ns.  Invoiced  at  75  ct.  per  yd.;  selling  price  $1.12g 

3.  The  duty  on  15  gross  London  porter,  allowing  10  % 
breakage,  was  $40.50,  at  20  °/0  ad  valorem:  how  much  a 
dozen  were  they  invoiced  at?  Ans.  $1.25 

ART.  296.    CASE  IV. — See  Case  IV,  Percentage. 

1.  French   cloths,  after   paying  30  %  duty,  and  other 
charges,  $73.80,  cost  in  store  $7389.03:  what  were  they 
invoiced  at?  Ans.  $5627.10 

2.  Imported  from  Havre  80  baskets  of  champagne,  12 
bottles  each,  5  %  breakage,  duty  40  %,  freight  and  other 
charges  $67.20,  and  the  whole  cost  $729.60:  what  did 
it  cost  a  bottle  at  Havre,  what  in  store,  and  how  much  a 
bottle  should  I  charge  to  clear  35  %  ? 

Ans.  50  ct.  at  Havre ;  80  ct.  in  store ;  selling  price  $1.08 

3.  The   cost  in    store  of  20   puncheons   Jamaica  rum, 
84  gal.    each,    is    $631.43;    duty    15  %,    leakage    2%, 
charges  $53.34  :  what  did  it  cost  a  gal.  in  Jamaica? 

Ans.  30  ct. 


XXIII.    INTEREST. 

ART.  297.  INTEREST  is  money  charged  for  the  use  of 
money. 

The  profits  accruing  at  regular  periods  on  permanent  inTestments, 
such  as  dividends  or  rents,  are  called  interest,  since  they  are  the 
growth  of  capital,  unaided  by  labor. 

R  B  v  i «  v. — 297.  What  is  interest  ?   What  arc  sometimes  called  interest  ? 


INTEREST.  243 


The  principal  is  the  sum  on  which  interest  is  charged. 

The  principal  is  either  a  sum  loaned;  money  invested  to  secure  an 
income;  or  a  debt,  which  not  being  paid  when  due,  is  allowed  by 
agreement  or  law  to  draw  interest. 

The  amount  is  the  principal,  with  its  interest  added. 

ART.  298.  Interest  is  payable  at  regular  intervals 
y  'arty,  half-yearly,  or  quarterly,  as  may  be  agreed  :  if  there 
is  no  agreement,  it  is  understood  to  be  yearly. 

ART.  299.  The  rate  of  interest  is  the  rate  per  cent,  of 
the  yearly  interest  to  the  principal :  it  is  called  rate  per 
cent,  per  annum;  per  annum  meaning  by  the  year,  in  Latin. 

If  interest  is  payable  half-yearly,  or  quarterly,  the  rate  is  still  the 
rate  per  annum,  or  rate  per  year.  In  short  loans,  the  rate  per  month 
is  generally  given ;  but  the  rate  per  year,  being  12  times  the  rate  per 
month,  is  easily  found :  thus,  2  f0  a  month  =  24.  fc  a  year. 

If  the  rate  of  interest  is  not  specified,  the  rate  establish- 
ed by  the  law  of  the  country  where  the  contract  is  made, 
prevails,  called  the  legal  rate,  which  is  generally,  but  not 
always,  the  highest  rate  allowed  by  the  law. 

If  a  higher  rate  of  interest  is  charged  than  the  law  al- 
lows, it  is  called  usury,  and  the  person  offending  is  subject 
to  a  penalty. 

ART.  300.  The  following  table  shows  the  legal  rate  in 
each  of  the  States  of  the  Union. 

LEGAL    RATES.  STATES. 

8  %  in  Georgia,  Alabama,  Mississippi  and  Florida. 
7  <f0  in  New  York,  So.  Carolina,  Michigan,  Wiscon.  and  Iowa. 
6  %  rn  Louisiana. 
10  %  in  Texas. 
6  %  in  all  other  places  in  the  U.  S.,  and  the  U.  S.  courts. 

ART.  301.     Interest  is  cither  Simple  or  Compound. 


REVIEW.  —  297.  What  is  tho  principal?  What  may  it  sometimes  be'. 
What  is  tho  amount?  293.  How  is  interest  payable?  If  there  is  no  agree- 
ment, how  is  it  payable?  299.  What  is  the  rate  of  interest?  For  what 
time  is  it  given  ?  When  is  it  given  per  month  ?  How  is  the  rate  per  yeai 
then  found?  If  no  rate  of  interest  is  mentioned,  what  one  prevail:-? 
What  is  usury  ?  301.  What  two  kinds  of  interest  ? 


244  RAY'S   HIGHER   ARITHMETIC. 

Simple  Interest  is  interest  which,  even  if  not  paid  when 
due,  is  not  convertible  into  principal,  and  therefore  can 
not  draw  interest  itself  and  accumulate  in  the  hands  of  the 
debtor,  however  long  it  may  be  retained. 

Compound  Interest  is  interest  which,  not  being  paid  when 
due,  is  convertible  into  principal,  and  from  that  time  draws 
interest  itself,  and  accumulates  in  the  hands  of  the  debtor, 
according  to  the  time  it  is  retained. 

Compound  interest  is  more  favorable  to  the  creditor 
than  simple  interest,  as  shown  in  this  example : 

If  I  lend  $100  at  the  rate  of  10  %  per  annum,  what 
should  I  receive  at  the  end  of  three  years,  presuming  no 
interest  paid  in  the  mean  time? 

SOLUTION  BY  SIMPLE  INTEREST. — The  interest  for  1  yr.  is  10  %  of 
$100  —  $10 ;  for  3  yr.  it  is  $30,  and  the  whole  sum  due  is  $130. 

SOLUTION  BY  COMPOUND  INTEREST. — The  interest  for  the  1st  year  is 
$10;  which,  not  being  paid,  makes  §110  then  due  and  drawing  in- 
terest: 10$  of  $110  —  §11,  the  interest  for  the  2d  year;  which, 
not  being  paid,  makes  $121  then  due  and  drawing  interest :  10  ^  of 
$121  =  $12.10,  the  interest  for  the  3d  year;  which  makes  the  whole 
Bum  then  due,  $133.10,  being  $3.10  more  than  by  Simple  Interest. 


XXIV.    SIMPLE  INTEREST. 

ART.  302.  SIMPLE  INTEREST  differs  from  the  other 
applications  of  Percentage  by  taking  the  time  into  con- 
sideration, which  they  do  not. 

All  questions  in  Percentage  involve  three  different  quan- 
tities, and  may  be  solved  by  a  simple  proportion.  All 
questions  in  interest  contain  four  different  quantities,  and 
may  be  solved  by  a  compound  proportion. 

ART.  303.  The  four  quantities  embraced  in  every  ques- 
tion of  interest  are,  1st,  the  principal;  2d,  ths  interest; 

REVIEW. — 301.  What  is  simple  interest?  Compound  interest?  Which 
U  more  favorable  to  the  creditor  ?  Show  the  difference  by  the  example. 
802.  How  docs  simple  interest  differ  from  the  other  applications  of  percent- 
age? If  the  time  were  left  oat  of  view,  how  could  questions  in  simple 
Interest  be  solved?  How  may  they  be  solved  as  it  is?  303.  What  four 
quantities  are  involved  in  every  question  of  simple  interest? 


SIMPLE   INTEREST. 


245 


3d,   the   rale ;    4th,   the   time :    of  these    four,   the  princi- 
pal being  the  standard  of  comparison,  is  100  per  cent. 

Any  three  of  these  quantities  being  given,  find  the  4th 
by  this 

GENERAL  RULE. 

Represent  the  principal  by  100  per  cent.,  and  proceed  ae  i* 
questions  of  compound  proportion. 

NOTE. — If  the  amount  is  given  or  required,  it  may  be  necessary  to 
perform  an  addition  or  subtraction  before  applying  the  rule,  or  after 
tho  result  has  been  obtained. 

Though  the  general  rule  is  sufficient  for  all  the  ordinary 
problems  in  Simple  Interest,  it  is  too  general  for  practical 
purposes,  and  a  special  rule  will  be  given  for  each  case. 

CASE   I. 

ART.  304.  Given,  the  principal,  rate  of  interest,  and 
time,  to  find  the  interest. 

RULE. — Find  the  yearly  interest  by  Case  I  of  Percentage; 
then  ascertain  from  it  the  interest  for  the  given  time  by  aliquot 
parts. 

NOTE. — In  calculations  of  interest,  every  month  is  regarded  as  30 
days,  and  every  year  as  12  months,  or  360  days. 


What  is  the  simple  interest  of  $354 
19  da.  at  6  %  ? 

SOL.  —  The  yearly  interest, 
being  C  f0  of  the  principal,  is 
found,  by  Case  I  of  Percentage; 
the  interest  for  3  yr.  7  mon. 
19  da.  is  then  obtained  by  ali- 
quot parts;  each  item  of  interest 
is  carried  no  lower  than  mills, 
<he  next  figure  being  neglected 
if  less  than  6;  but  if  6  or  over 
it  is  counted  1  mill. 


80  for  3  yr.  7  mon. 


lyr.  = 

$354.80 
6 

21.2880 

3yr. 
6  mon. 
1  mon. 
18  da. 
Ida. 

^_  o 

=  * 
=  k 

=  T>* 

=  a'o- 

63.864 
10.644 
1.774 
1.064 
.059 

$77.41  Am 


RK VIEW. — 303.  How  many  must  be  known  ?  What  is  the  general  rula 
for  questions  of  simple  interest?  If  tho  amount  is  given  or  required,  what 
maybe  necessary?  304.  What  is  Cose  1?  Tho  rule?  How  many  days 
are  counted  a  month  in  interest  ?  How  many  a  year  ?  Solve  the  example. 
What  custom  prevails  in  computing  interest?  Why? 


240  RAY'S   HIGHER  ARITHMETIC. 


If  the  rate  of  interest  were  5  %,  7j  %,  10  %,  &c.  the  pro- 
cess would  be  similar ;  but,  as  the  prevailing  rate  is  cither 
6  %,  or  a  simple  aliquot  part  greater  or  less  than  6  %,  it  is 
customary,  whatever  be  the  rate,  to  compute  the  interest,  first  at 
6  %  ;  then  increase  or  diminish  the  result  by  such  a  part 
of  itself,  as  may  be  necessary  to  obtain  the  interest  at  the  ratt 
given. 

A   SHORT   METHOD. 

ART.  305.  As  6  %  a  year  is  the  same  as  1  %  for  '2 
months,  take  1  %  of  the  principal,  by  pointing  off  the  two 
right  hand  figures  of  dollars;  the  result  is  the  interest  for 
2  mon.,  or  60  da.;  and  the  interest  for  the  given  time  can 
be  found  by  aliquot  parts. 

Applying  this  method  to  the  last  example,  the  operation 
will  appear  thus:  R  q  54  80 


SOLUTION. — Write  the  principal,  $354.80 ;  cut        7  0 
off  the  2  right  hand  figures  of  dollars  by  a  vertical  \ 

line ;  $3.548  is  the  interest  for  2  mon. :  $70.90  is  1 

the  int.  for  40  mon.,  being  20  times  the  int.  for 
2  mon.  just  above;  $1.774  is  the  int.  for  1  mon., 


96 
774 
064 
059 


41 


being  ^  the  top  number;  $1.064  is  the  int.  for 
18  da.,  bein  T30  of  the  top  number;  (see  Rcm.,  Art.  234),  and  the  $.059 
is  the  int.  for  1  day,  being  3*5  of  $1.774,  the  int.  for  1  mon.     The 
sum  of  these  is  the  int.  for  43  mon.  19  da.  =  3  yr.  7  mon.  19  da. 

ART.  306.  After  finding  the  Int.  at  6%,  observe,  that  the 
Interest  at  5  ^  =  interest  at  6  ft,  —  ^  of  itself.     And  the  Interest 


at  1    %  =  \  Int.  at  Q%. 
at  7    #=Int.atG<6-f-£  of  it. 
at7A#  =     "      6  #-f- £  of  it 
at  8    %=     "      Sfc+l  of  it. 
at  9    %=    « 


at  4^  =  Int.  at  6^  —  ^of  it. 
at  4    ^  =:  Int.  at6  $ — \  of  it. 
at  3    %  =  3  Int.  at  6  %. 
at  2    %  =  \  Int.  at  6  %. 
at  1^  %  =|  Int.  atG  %. 
at  12  %,   18  %,   24  %  —  2,  3,  4  times  Interest  at  6  %. 
at    5  %,    10  %}    15  %,   20  %  =  T2-    g.   ?i    3    Interest  at  6  % 
after  moving  the  point  one  figure  to  the  right. 

ART.  307.     ANOTHER  METHOD.    Take  the  example  al 
ready  solved. 

REVIE  w.  —  305.  Explain  the  short  method.  306.  After  the  interest  at 
6  %  has  been  found,  how  is  the  interest  at  5  ^  obtained?  At  4J  f0  ? 
4  %,  Ac.  7  %  ?  ?i  %,  Ac-  307.  Explain  "another"  method. 


SIMPLE  INTEREST. 


247 


$354.80 


$177.40 
.436j 

70960 
638640 
5913 


$77.40553 
$77.41  Am. 

$17.740 


7096 
639 


.436J 


ANALYSIS. — The  interest  of  any  sum 
($354.80)  at  6  %  equals  the  interest  of  half 
that  sum  ($177.40)  at  12  %.  But  12  %  a 
year  equals  1  %  a  mon.,  and  for  3  yr.  7  mon. 
=  43  mon.,  the  rate  is  43  <&,  and  for  19  da., 
which  is  -]  §  of  a  mon.,  the  rate  is  ]  $  %  ', 
hence,  the  rate  for  the  whole  time  is  43]  §  %, 
and  43]  \  %  of  the  principal  will  be  the 
interest.  To  get  43]  \%,  multiply  by  43] § 
hundredths  =  .43] g  «=  .436]  (Art.  147). 

By  using  the  contracted  multiplication 
(explained  in  Art.  153),  the  work  may  be 
shortened,  and  the  answer  obtained  correct- 
ly to  cents.  t> 

$77.41 

In  the  multiplier  .436s,  the  hundredths  (43)  are  the 
number  of  months,  and  the  thousandths  (6])  are  3  of  the 
days  (19  da.),  in  the  given  time,  3yr.  7  mon.  19  da. 

TWO   PRACTICAL    RULES. 

RULE  1. — Express  the  years,  if  any,  in  months,  and  write  the 
whole  mimber  of  months  as  decimal  hundredlhs ;  after  which,  place 
\  of  the  days,  if  any,  as  thousandths:  multiply  half  the  principal, 
by  this  number. 

RULE  2. —  Or,  point  off  from  the  pnncipal  two  more  decimals 
than  it  already  has.  This  gices  the  interest  for  2  months,  or  GO 
days,  from  wlrich  the  interest  for  the  given  time  can  be  found  by 
aliquot  parts. 

Either  of  these  mles  gives  the  interest  at  G  %,  from  which  tht 
interest  at  any  other  rate  can  be  found  by  aliquot  parts. 

NOTE. — In  the  1st  rule,  when  the  number  of  days  is  1  or  2,  place 
a  cipher  to  the  right  of  the  months,  and  write  the  £  or  f  :  otherwise, 
they  will  not  stand  in  the  thousandths'  place :  thus,  if  the  time  is 
I  yr.  4  mon.  1  da.,  the  multiplier  is  .IGO]  ;  for  9  mon.  2  da.,  .090|. 

REV  uw. — 307.  How  can  tho  work  bo  shortened?  What  are  tho  hun- 
Jredths  of  tho  multiplier  equal  to?  Tho  thousandths?  What  is  tho  first 
rule  for  computing  interest?  The  2il  rule?  In  using  th«  1st  rule,  when 
the  days  are  1  or  2,  what  is  necessary?  When  tho  fraction  in  tho  multi- 
plier is  8,  how  d<5  we  proceed?  Which  rule  is  generally  most  cou/euient 
for  short  times  ?  In  taking  aliquot  parts  for  tho  days  in  the  2d  rule,  what 
ihould  we  make  use  of? 


248  RAY'S   HIGHER   ARITHMETIC. 


If  the  fraction  in  the  multiplier  is  f ,  it  ia  better  to  take  3  of  the  tohoU 
principal,  than  3  of  /ia//the  principal. 

REMARK. — The  2d  rule  will  generally  be  found  most  convenient 
for  short  times,  and  will  not  require  more  than  two  aliquot  parts 
for  the  days,  by  using  the  tenths,  as  well  as  the  halves,  fourths,  &c. 
(See  Art.  234,  Hem.) 

A.RT.  308.  In  New  York  and  some  other  parts  of  the 
II.  S.,  and  in  Great  Britain,  the  interest  for  days  is  calcu- 
lated at  365  days  to  the  year.  In  those  places,  1  day's 
interest  =  TJSS  of  a  year's  interest  =  355  of  360  da.  i-- 
terest  obtained  by  the  rule  s  iff  or  7;i  of  a  day's  interest, 
as  commonly  obtained;  but  |$  is  Vs  less  than  the  whole: 

Hence,  the  interest  for  any  number  of  days,  counting  365 
da.  to  a  year,  is  73  few  than  the  interest  for  the  same  number 
of  days,  counting  360  da.  to  a  year. 

FIND  THE  SIMPLE  INTEREST  OP  AMg 

1.  $1*78.63  for  2yr.  5mon.  26da.,  at. 7  %.  =  $31.12 

2.  $6084.25  for  lyr.  3 num.,  at  4  A  %.    =$342.24 

3.  $64. 30  for  Iyr.l0mon.l4da.,at  9%.=  $10.83 

4.  $1052. 80  for  28  da.,  at  10%.  =$8.19 

5.  $419.10  for  8  mon.  16  da.,  at  6%.        =$17.88 

6.  $1461.85for6yr.7mon.4da.,atlO%.=$904.01 

7.  $2601.50  for  72  da.,  at  *l$%.  =  $39.02 

REMARK. —  72  da.  =  2  mon.  12  da.      * 

8.  $8722.43  for  5Ayr.,  at  6%.  =  $2878.40 

9.  $326.50  for  1  mon.  8  da.,  at  8  %.         =  $2.76 

10.  $1106.70  for  4yr.  Inion.  Ida.,  at  6%.  =  $271.33 

11.  $10000  for  Ida.,  at  Q%.  =  $1.07 

12.  $4642. 68  for  5 mon.  I7da.,  at  15  %.  =  $323.05 

13.  $13024  for  9mon.  13da.,  at  10%.  =  $1023.83 

14.  $615.38for4yr.llmon.6da.,at20%.=  $607.l7 

15.  $2066.19for3yr.6mon.2da.,at30%.=$2l72.94 

16.  $92.55  for  3  mon.  22  da.,  at  b%.  =  $1.44 


REVIEW. — 307.  Into  how  many  aliquot  parts  can  tho  days  always  be 
dlrided?  308.  What  is  said  of  interest  for  days  in  New  York,  Great 
Britain,  Ac.?  How  is  interest  for  any  number  of  days  obtained  in  those 
places?  Why? 


SIMPLE   INTEREST 


Find  the  simple  interest  of  ANiJ 

17.  $1532. 45  for  9 yr.  2  mon.  7  da.,  at  12%. 

=  $1689.27 

18.  $78084.50  for  2yr.  4  mon.  29  da.,  at  IS%. 

=  S33927  72 

19.  $512.60  for  8mon.  18da.,  at  7  %.       =  $25*.72 

20.  $3278.12forlyr.6mon.3da.,at4%.=  $197.78 

21.  $8408. 46  for  11  mon.  5 da.,  at  3%.  =$234.74 

22.  $126 . 75  for  2  yr.  24  da.,  at  8  % .          =$20.96 

23.  $1363.20  for  39  da.,  at  11  %  a  month.  =  $22.15 

24.  $402. 50  for  100  da.,  at  2  ^  a  month.  =$26.83 

25.  $6919.32  for  7yr.  6 mon.,  at  6  %.    =  $3113.69 

26.  $990.73  for  9  mon.  19  da.,  at  7  % .         =$55.67 
Find  the  amount  of 

27.  $757.35  for  117  da.,  at  H  %  a  mon.  =  $801.65 

28.  $1 883  for  1  yr.  4 mon.  21  da.,  at  6  % .  =  $2040 .23 

29.  $5000forlOyr.lOmon.20da.,at9$.  =  $9900. 

30.  $4212.45for  5yr.5mon.25da.,at5%.=$5367.95 

31.  $262.70  for  53da.  at  I  %  a.  month.    =  $267.34 

32.  $584.48  for  133 da.,  at  7£  %.  =  $600.67 

33.  $8291.56  for  294 da.,  at  6%.          =$8697.85 

34.  $16372.05  for  3yr.  9mon.,at  8  %.=  $21283.67 

35.  $2001. 25  for  86 da.,  at  B%.  =$2029.93 

36.  $392.28  for  71  da.,  at  2A  %  a  mon.    =  $415.49 

37.  $3032. 90  for  7mon.  7 da.,  at  1  %.    =$3160.87 

38.  Find  the  interest  of  $7302.85  for  365  da.,  at  6  %t 
counting  360  da.  to  a  year.  Ans.  $444.26 

39.  Of  $10000  for  360  da.,  at  6  %,  counting  365  days 
to  a  year.     (Art.  308.)  Ans.  $591.78 

40.  Same  «is  last,  360  da.  to  a  yr.  Ans.  $600. 

41  If  I  borrow  $1000000  in  New  York,  at  7  %,  and 
lend  it  at  7  %  in  Ohio,  what  do  I  gain  in  180  da.? 

Am.  $479.45 

42.  Find  the  interest  of  $5064.30  for  7mon.  12  da.,  at 
7  %,  in  New  York.  (Art.  308).  Ans.  $218.45 

SUGGESTION. — Find  the  interest  for  years  and  months  in  New 
York  as  elsewhere;  but  for  days,  find  the  interest  as  usual,  and 
diminish  it  by  7's  of  itself,  before  adding  it  in. 


RAY'S   HIGHER   ARITHMETIC. 


Find  the  simple  interest  of 

43.  $681.75  for  98 da.,  1%  in  N.  Y.  Ans.  $12.81 

44.  $1353.10  for  2yr.  8  mon.  29  da.,  at  7  %  in  New 
York.  Ans.  $260.10 

45.  $6786. 24  for  lyr.  10  mon.  16  da.,  at  10^  in  New 
York.  Ans.  $1273.89 

46.  If  I  borrow  $12500  at  6  %,  and  lend  it  at  10  %, 
what  do  I  gain  in  3  yr.  4  mon.  4  da.?     Ans.  $1672.22 

47.  If  I  borrow  $23275  at  12  %,  and  invest  it  at  7  %, 
what  do  I  lose  in  2  yr.  1  mon.  23  da.?      Ans.  $2498.83 

48.  What  is  the  interest  of  $3416.20,  at  6  %,  from 
Feb.  3,  1847,  to  Aug.  9,  1851?  Ans.  $925.79 

REMARK. — Find  the  time  by  Art.  325,  Rem.  3. 

49.  If  $4603.15  is  loaned  July  17,  1853,  at  7  % 
what  is  due  March  8,  1855?  Ans.  $5130.34 

50.  Find  the  interest  at  8%  of  $13682.45,  borrowed 
from  a  minor  13  yr.  2  mon.  10  da.  old,  and  retained  till 
he  is  of  age  (21  yr).  Ans.  $8543.93 

51.  What  is  the  amount  of  $5772  from  Oct.  26,  1850, 
to  April  12,  1855,  at  10%?  Ans.  $8348.56 

52.  In  one  year,  a  broker  loans  $876459.50  for  63  da., 
at  1$  %  a  mon.,  and  pays  6  %  on  $106525.20  deposits: 
what  is  his  p;ain?  Ans.  $21216.96 

53.  What  is  a  broker's  gain  in  1  yr.,  on  $100  deposited 
at  6  %,  and  loaned  11  times  for  33  da.  at  2  %  a  mon.? 

Ans.  $18.20 

54.  Fird  the  simple  interest  of  £493  16s.  8d.  for  lyr. 
8  mon.  at  6  %.  Ans.  £49  7s.  8d. 

SUGGESTION. — Find  the  interest  of  sterling  money  by  Rule  in 
Art.  304,  multiplying  and  dividing  as  in  compound  numbers;  or, 
reduce  the  principal  to  one  denomination,  as  pounds',  and  apply  a 
Rule  i*  Art.  307. 

55.  Find  the  simple  interest  of  £24  18s.  9d.  for  10  mon. 
»»  6  %.  Ans.  £1  4s.  Hid. 

56.  Of  £25  for  1  yr.  9  mon.  at  5  %.    Ans.  £2  3s.  9d. 

57.  Of  £651  for  7  mon.  at  4A  %.    Ans.  £17  Is.  9jd. 
68.    Of  £648  15s.  6d.  from  June  2  to  November  25 

at  5  %.     (Art.  308.)  Ans.  £15  12s.  lOd. 

§9.    Of  £14  from  March  23  to  Nov.  2,  at  6  %. 

Ans.  10s.  3|d. 


SIMPLE   INTEREST.  251 

60.  Find  the  simple  interest  of  £66  8s.  from  May  6  to 
Aug.  21,  at  bk%.  Ans.  £1  Is.  5d. 

61.  Of£98for3yr.l22da.at6$.  Am.  £19  12s. lid. 

62.  Of  £374  5s.  from  April  1  to  Dec.  29,  at  4  %» 

Ans.  £11  3s.  1R 
CASE  II. 

ART.  309.  Given,  the  principal,  interest,  and  time,  to 
find  the  rate. 

RULE. — Assume  1  per  cent,  for  the  rate;  determine  the  interest 
on  this  supposition,  and  divide  the  given  interest  by  it. 

PROOF. — Calculate  the  interest  at  the  rate  thus  found; 
if  it  agrees  with  the  given  interest,  the  work  is  right. 

NOTE. —  If  the  amount  ig  given  instead  of  the  interest,  the 
principal  may  be  subtracted  from  it,  to  obtain  tLe  interest. 

If  I  loan  $6875  at  simple  interest,  and  in  2yr.  3  mon. 
18 da.  receive  $7823.75,  what  is  the  rate? 

SOLUTION. — 'Assume  1  %  for  the  rate;  the  interest  of  $6875  for 
2yr.  3 mon.  18  da.,  at  1  %,  is  $158.125,  (Case  I);  the  given  interest 
=  $7823.75  —  $6875  =  $948.75,  and  since  this  contains  $158.125 
six  times,  it  must  have  accumulated  at  a  rate  six  times  as  great; 
that  is,  6  fe. 

1.  What  is  the  rate  of  interest,  when  I  pay  $119.70 
for  the  use  of  $3325  for  10  mon.  24  da.  ?       Ans.  4  %. 

2.  If  $65.47  be  paid  for  a  loan  of  $844.75  for  93 
days,  what  is  the  rate  per  month?  Ans.  2i  %. 

Suo. — Find  the  rate  per  year  by  the  rule,  and  divide  it  by  12. 

3.  If  I  borrow  $5000  for  7  yr.   6  mon.   28  da.,   and 
return  $10000,  what  is  the  rate?  Ans.  13&7T  %- 

4.  At  what  rate  per  annum  will  any  sum  of  money 
double  itself  by  simple  interest,  in  5,  6,  7,  8,  9,  10,  12, 
15,  20,  25  years,  respectively? 

AM.  20,  161,  144,  121,  1U,  10,  8i,  6§  5,  4#. 

SUGGESTION. — Take  $100  for  the  principal,  and  $100  for  the  in- 
wrest.  If  100  be  divided  by  the  time  in  years,  the  quotient  will  be  the  rate 
at  which  any  turn  will  double  itself  at  simple  interest  in  that  time. 


REVIEW.  — 309.  What  is  Case  2?  The  rule?  The  proof?  If  the 
amount  it  given  instead  of  the  interest,  what  is  necessary?  Solve  th« 
•zample. 


252  RAY'S   HIGHER   ARITHMETIC. 

5.  At  -what  rate  per  annum  will  any  sum  treble  itself 
at  simple  interest,  in  5,  10,  15,  20,  25,  30  years,  respec- 
tively? Ans.  40,  20,  13i,  10,  8,  6ii  %• 

6.  At  what  rate  per  annum  will  any  sum  quadruple 
itself  at  simple  interest,  in  6,  12,  18,  24,  30  years,   re- 
spectively? Ans.  50,  25,  161,  12-i,  10  %. 

7.  What  is  the  rate  of  interest,  when  $35000  yieldi 
AH  income  of  $175  a  month?  Ans.  6  %. 

8.  When  $29200  produces  $6.40  a  day?    Ans.  8  %. 

9.  When    $12624.80  draws  $315.62   interest  quar- 
terly?  Ans.  10  %. 

10.  When  stock  bought  at  40  %  discount,  yields  a  semi- 
annual  dividend  of  5  %  ?  Ans.  16f  <f0  per  annum. 

11.  A  house  that  cost  $8250,  rents  for  $750  a  year; 
the  insurance  is  -fo%,  and  the  repairs  3  %,  every  year' 
what  rate  of  interest  does  it  pay?  Ans.  8  % — . 

CASE    III. 

ART.  310.  Given,  the  interest,  rate,  and  time,  to  find 
the  principal. 

RULE. — Assume  $1  for  the  principal;  determine  (he  interest 
on  this  supposition,  and  divide  the  given  interest  by.it. 

PROOF. — Calculate  the  interest  on  the  principal  found; 
if  it  agrees  with  the  given  interest,  the  work  is  right. 

NOTES. — 1.  After  the  principal  has  been  found,  the  interest  may 
be  added  to  it,  and  the  amount  thus  obtained. 

2.  The  contracted  method  of  division  in  Art.  168  may  be  generally 
used. 

What  sum  will  yield  $228.80  interest  in  4mon.  23 da., 
at  15  %  per  annum? 

SOLUTION. — Assume  $1  for  the  principal;  the  interest  of  it  for 
4  mon.  23  da.,  at  15  %  is  $.059T73  (Case  I) :  as  the  given  interest, 
$228.80,  contains  this  3840  times,  the  principal  producing  it  must 
XJ  3840  times  as  large,  that  is,  $3840. 

WHAT   PRINCIPAL   WILL   PRODUCE 

1.  $1500  a  year,  at  6  %  ?  Ans.  $25000. 

2.  $1830  in  2  yr.  6  mon.,  at  5  %  ?      Ans.  $14640. 

REVIEW. —  310.  What  is  Case  3?  The  rule  ?  The  proof?  How  can 
the  amount  be  afterward  found  ? 


SIMPLE   INTEREST.  253 


What  principal  will  produce 

3.  $45  a  mon.,  at  9  %  ?  Am.  $6000. 

4.  $17  in  68  da.,  at'l  %  a  mon.?  Ans.  $750. 
5     $86.15in9mon.llda.,atlO^?    Ant.  $1103.70 

6.  $313.24  in  112  da.  at  7  %  ?      Ant.  $14383.47 

7.  $146.05in7mon.  14 da.  at  Q%1  Ans.  $3912.05 

8.  $79.12  in  5  mon.  25  da.  at  7  %  in  N.  Y.? 

Ans.  $2329.72 
CASE  IV. 

ART.  311.  Given,  the  amount,  rate,  and  time,  to  find 
the  principal. 

RULE. — Assume  $1  for  the  principal;  determine  the  amount 
on  that  supposition,  and  divide  the  given  amount  by  it. 

PROOF. — Calculate  the  amount  on  the  principal  found; 
if  the  same  as  the  given  amount,  the  work  is  right. 

NOTES. — 1.  After  finding  the  principal,  subtract  it  from  th« 
amount,  to  get  the  interest. 

2.  The  contracted  method  of  division  (Art.  158)  may  be  used. 

What  sum,  drawing  simple  interest  at  5%,  will  amount 
to  $819.45  in  1  yr.  8mon.  5  da.? 

SOLUTION. — Assume  $1  for  the  principal;  the  amount  of  $1  for 
lyr.  8mon.  5da.,  at  6  ^,  is  $1.0843-1g  (Case  I);  as  the  given  amount 
$819.45,  contains  this  755.93  times,  it  must  have  arisen  from  a  prin- 
cipal 765.93  times  as  large ;  that  is,  $755.93. 

1.  What  principal  in  2  yr.  3  mon.  12  da.,  at  Q%,  will 
amount  to  $1367.84?  Ans.  $1203.03 

2.  What  principal  in  10  mon.  26  da.,  will   amount  to 
$2718.96,  at  10  %  interest?  Ans.  $2493.19 

3.  What  principal,  at  4^%,  will  amount  to  $4613.36 
in  3  yr.  1  mon.  7  da.  ?  Ans.  $4048.14 

4.  What  principal,  at  7%,  will  amount  to  $562.07  in 
79  da.  (365  da.  to  a  year)  ?  Ans.  $553.68 

PRESENT    WORTH 
ART.  312.     Is  an  important  application  of  Case  IV. 

REVIEW.  —  310.  What  method  of  division  may  be  generally  used? 
Solve  the  example.  311.  What  is  Case  4?  The  rule?  The  proof  1 
How  is  the  interest  afterward  obtained?  What  method  of  division  may  be 
nsod  ?  Solve  the  example. 


854  RAY'S  HIGHER   ARITHMETIC. 


Present  Worth  is  a  phrase  used  in  speaking  of  a  debt 
before  it  is  due,  and  is  the  sum  which  at  the  prevailing  rate 
of  interest,  will  amount  to  that  debt  wlwn  it  is  due. 

The  difference  between  the  present  worth  of  a  debt  and 
the  debt  itself,  is  the  interest  of  the  present  worth  from  the 
present  time  until  the  time  the  debt  is  due;  it  is  called 
the  discount,  since  it  is  the  sum  which  must  be  deducted 
from  the  debt  or  nominal  value,  to  give  the  present  worth  or 
real  value. 

REMARK. — The  discount  in  this  case,  like  that  in  Art.  27-5,  is  an 
abatement  or  deduction  from  the  apparent  value,-  in  fact,  the  difference 
between  the  real  and  tlie  nominal  value:  but  the  latter  is  a  certain  per 
cent,  of  the  nominal  value,  \vhile  this  is  a  certain  per  cent.,  not  of  the 
nominal  value  (the  debt),  but  of  the  real  value  (the  Present  Worth). 

1.  Find  the  present  worth  and  discount  of  $5101.75 
due  in  1  yr.  9  mon.  19  da.,  rate  of  interest,  6  %. 

Ans.  Pros,  wor.,  §4603.775  ;  Dis.,  $497.975 

2.  Also,  of  $1476.81,  due  in  4 mon.  11  da.,  rate,  Q%. 

Ans.  Pres.  wor.,  $1445.26;  Dis.,  $31.55 

3.  Find  the  present  worth  of  $2906.30,  due  in  103 
days,  rate,  8  %.  Ans.  $2841.27 

4.  Find  the  discount  of  $6344.25,  due  in  23  days,  rate 
of  interest,  5  %.  Ans.  $20.20 

5.  Find  the  present  worth  of  $12720.40,  due  in  9  da., 
at  7  %  in  New  York.  Am.  $12698.48 

6.  I  can  sell  property  for  $7500   cash,  or  for  $4250 
payable   in    6  mon.   and  $4000    payable   in    1  yr. :    which 
should  I  prefer?  and  what  do  I  gain   by  it  if  money  is 
worth  12  %  to  me?  Ans.  The  latter;   $80.86 

CASE  V. 

ART.  313.  Given,  the  principal,  rate,  and  interest,  to 
find  the  time. 

RULE. — Assume  1  year  for  the  time;  determine  the  interest  on 
ttds  supposition,  and  divide  the  given  interest  by  it. 

PROOF. — Calculate  the  interest  for  the  time  thus  found 
if  it  is  the  same  as  the  given  interest,  the  work  is  right. 

REVIEW.— 312.  Why  is  this  case  important?  What  is  present  worth  T 
What  is  the  difference  between  a  debt  not  duo  and  its  present  worth  ? 
What  is  it  called  ?  Why?  313.  What  is  Case  6?  The  rule?  The  proof  T 


SIMPLE  INTEREST.  355 


NOTES. — 1.  The  quotient  will  be  the  time  in  years;  if  it  contain 
i  fraction,  reduce  it  to  months  and  days  by  Art.  222. 

2.  If  the  amount  is  given,  subtract  the  principal  from  it,  to  get 
the  interest;  then,  apply  the  rule. 

In  what  time  will  $830  amount  to  $1000,  at  6  %  simple 
interest? 

SOLUTION. — Assume  1  yr.  for  the  time;  the  interest  if  §830  for 
lyr.  at  6  %  is  $49.80  (Case  I) :  the  given  interest  =  $1000  —  $830  =a 
$170,  and  as  this  contains  $49.80,  3^|  times,  it  must  have  taken 
83 J|  times  as  long  to  accumulate,  that  is,  S^ffyr.  =  Syr.  4  mon, 
29  da.,  by  reduction.  (Art.  222.) 

IN   WHAT   TIME   WILL 

1.  §1200  amount  to  $1800,  at  10  %  ?         An*.  5  yr. 

2.  $415. 50  to  $470. 90,  at  10  %?    Ant.  lyr.  4mon. 

3.  $3703.92  to  $4122. 15,  at  8%? 

Ans.  1  yr.  4  mon.  28  da. 

NOTE. — A  part  of  a  day  is  omitted  in  the  answer,  not  being  re- 
cognized in  interest ;  it  must  be  taken  account  of,  however,  in  the 
proof. 

•  4.  In  what  time  will  any  sum,  as  $100,  double  itself  by 
simple  interest  at  4£,  5,  6,  7,  7s,  8,  9,  10,  12,  12£,  15, 
18,  20,  25,  30  %  ? 

An*.  22|,  20,  161,  14?,  13i,  12*,  11$  10,  8i,  8,  6|, 
5§,  5,  4r  83  yr.  100  divided  by  the  rate  of  interest,  will  give 
the  number  of  years  in  which  any  sum  will  double  itself. 

5.  In  what  time  will   any  sum  treble  itself  by  simple 
interest  at  4,  4i,  5,  6,  7,  7i  8,  9.  10,  12  %  ? 

Ant.  50,  44|,  40,  33|,  284,  26f,  25,  22§,  20,  IGfyr. 

6.  In  what  time  will  any  sum  quadruple  itself  by  simple 
Int.,  at  5,  6,  7,  8,  10,  12,  15,  20,  30,  40,  50,  100%? 

Ant.  60,  50,  42$,  371,  30,  25,  20, 15,  10,  7i  6,  3  yr. 

7.  How  long   must   I   keep  on   deposit   $1374.50,  at 
10  %,  to  pay  a  debt  of  $1480.78?       Ant.  9  mon.  8  da. 

8.  How  long  will  it    take    $3642.08    to   amount    to 
$4007.54,  at  12  %  ?  Ans.  10  mon.  1  da. 

9.  How  long  would  it  take  $175.12  to  produce  $6.43 
interest  at  6  %1  Ans.  7  mon.  10  da. 

REVIEW. — 313.  What  will  the  quotient  be?  If  it  contains  a  fraction, 
what  is  necessary  ?  What,  if  the  amount  ia  given  ?  Solve  the  example. 
What  is  said  of  a  part  of  a  day  ? 


250  RAY'S  HIGHER   ARITHMETIC. 


10.  How  long  would  it  take  $415.38  to  produce 
$10.09  interest  in  New  York  at  7  %t  Am.  134  da. 

REMARK. — In  this  example,  multiply  the  fraction  of  the  year  by 
305,  instead  of  12  and  30.  (Art.  308.) 

ART.  314.  All  the  Cases  of  Simple  Interest,  except  the 
4th,  which  treats  of  Present  Worth,  can  be  solved  by 
Compound  Proportion,  after  the  following  form: 

$100.    Principal. 

:  :    :     Rate.  :     Interest. 

1  yr.      Time  in  yr. 

ART.  315.  The  last  4  cases  show  a  striking  similarity  in 
their  rules  and  modes  of  operation,  and  can  be  put  into  a 

GENERAL    RULE   FOR   THE   LAST    FOUR   CASES. 

Assume  1  for  the  quantity  required;  determine  tlie  interest  (or 
Amount  if  it  be  necessary),  on  this  supposition,  and  divide  the 
jicen  interest  (or  amount)  by  it. 


XXV.   BANKING. 

ART.  316.  The  most  important  application  of  Simple 
Interest  is  Banking,  including  the  Discounting  of  Notes, 
Exchange,  and  the  settlement  of  depositors'  accounts. 

BANKS  are  corporations  which  deal  in  money. 

A  bank  of  issue  is  one  which  issues  its  own  notes  as 
money.  A  bank  of  discount  is  one  which  makes  loans. 
A  bank  of  deposit  is  one  which  takes  charge  of  money 
belonging  to  others. 

REMARK. — A  bank  is  generally  controlled  by  a  board  of  direc- 
tors, elected  by  the  stockholders ;  the  principal  officers  are  the 
president  and  cashier. 

REVIEW. — 314.  By  what  compound  proportion  may  all  the  cases  of 
interest,  except  the  4th,  bo  solved  ?  Why  can  not  the  4th  be  solved  thus  ? 
315.  What  general  rule  serves  for  the  last  4  cases?  In  which  case  only 
will  it  be  necessary  to  use  the  amount?  Why?  316.  What  is  the  most 
important  application  of  simple  interest?  What  does  it  include?  What 
are  banks  ?  What  kinds  ?  Define  each.  How  is  a  bank  generally  con- 
troll  -d  ?  What  are  the  principal  officers  ? 


BANKING.  257 


PROMISSORY  NOTES. 

ART.  317.  A  PROMISSORY  NOTE  is  a  written  promise  bj 
one  party  to  pay  a  named  sum  to  another. 

The  person  by  whom  the  note  is  signed  is  the  maker 
of  the  note.  The  person  to  whom  the  money  is  promised 
is  the  payee.  The  owner  of  a  note  is  the  holder. 

A  promissory  note  is  negotiable,  when  it  is  payable  to 
bearer,  or  to  the  order  of  the  payee ;  otherwise,  it  is  not 
negotiable. 

A  negotiable  note  may  pass  from  hand  to  hand ;  the 
payee  or  holder  endorsing  it  by  writing  his  name  on  its 
back,  thus  becoming  liable  for  its  payment,  if  the  note  is 
payable  "  to  order." 

If  the  note  is  payable  "  to  bearer,"  no  indorsement  is 
required  on  transferring  it,  and  only  the  maker  is  re- 
sponsible. 

It  is  essential  to  a  valid  promissory  note,  that  it  contain 
the  words  "value  received,"  and  that  the  sum  of  money 
to  be  paid  should  be  written  in  words. 

The  face  of  a  note  is  the  sum  promised  to  be  paid. 

If  a  note  contain  the  words  "with  interest,"  it  drawa 
Interest  from  date,  and  if  no  rate  is  mentioned,  the  legal 
rate  prevails. 

The  face  of  such  notes  is  the  sum  mentioned  with  ita 
interest  from  date  to  the  day  of  payment. 

If  a  note  does  not  contain  the  words  "with  interest," 
and  is  not  paid  when  due,  it  draws  interest  from  the  day 
of  maturity,  at  the  legal  rate,  till  paid. 

REMARK. — The  day  of  maturity  is  when  the  note  is  legally  due. 

If  a  note  is  not  paid  at  maturity,  the  indorsers  must  be 
notified  of  the  fact,  in  writing,  when  it  falls  due,  or  they 
will  not  be  liable.  This  notification  is  generally  made  by 
a  Notary,  and  is  called  a  Protest. 

REVIEW. — 317.  What  is  a  promissory  note?  Who  is  tho  maker  of  a 
note  ?  The  payee  ?  When  is  a  promissory  note  negotiable  ?  What  may 
be  done  with  a  negotiable  note  ?  If  it  is  payable  to  order,  what  is  noces- 
lary?  What  is  indorsing?  What  effect  docs  it  have?  What  kind  of  n 
note  needs  no  indorsement  when  transferred  ?  What  are  essential  to  a 
ralid  promissory  note  ?  What  is  the  face  of  a  note  ?  What  is  tho  face, 
when  the  note  contains  the  words  "with  interest?" 

22 


RAY'S   HIGHER   ARITHMETIC. 


ART.  318.    If  a   note   is   payable  "on   demand,"  it  is 
legally  due  when  presented. 
Bank  notes  are  of  this  sort,  being  payable  "to  bearer  on  demand." 

If  a  day  of  payment  is  specified  in  the  note,  it  is  due 
the  3d  day  afterward;  in  some  places,  it  is  due  on  the 
day  specified. 

If  a  note  is  payable  a  certain  time  "  after  date,"  proceed 
thus, 

TO   FIND  THE   DAY  A  NOTE   IS    LEGALLY  DUE. 

RULE. — Add  to  the  date  of  the  note,  lite  number  of  years  and 
months  to  elapse  before  payment ;  if  this  gives  the  day  of  a  month 
higher  than  that  month  contains,  take  the  last  day  in  that  month; 
then,  count  the  number  of  days  mentioned  in  the  note  and  3  more : 
this  will  give  the  day  the  note  is  legally  due;  but  if  it  is  a  Sunday 
or  a  national  holiday,  it  must  be  paid  the  day  previous. 

NOTES. — 1.  When  counting  in  the  days,  do  not  reckon  the  one 
from  which  the  counting  begins.  The  three  additional  days  are 
called  "days  of  grace";  in  some  countries  they  are  4,  5,  or  more. 
The  day  before  "grace"  begins,  the  note  is  nominally  due;  it  is 
legally  due  on  the  last  day  of  grace. 

2.  The  months  mentioned  in  a  note  are  calendar  months.  Hence, 
a  3  mon.  note  would  run  longer  at  one  time  than  at  another ;  one 
dated  Jan.  1st,  will  run  93  days,  one  dated  Oct.  1st,  will  run  95 
days :  to  avoid  this  irregularity,  the  time  of  short  notes  is  generally 
given  in  days  instead  of  months;  as,  30,  GO,  90  days,  instead  of 
1,  2,  3  months. 


DISCOUNTING  NOTES. 

ART.  319.  DISCOUNTING  NOTES  is  buying  them  at  a 
discount,  and  is  chiefly  done  by  banks  and  brokers. 

Notes  to  be  discounted  must  be  payable  to  order,  and  indorsed  by 
the  payee. 

The  proceeds  or  cost  of  a  note  is  the  sum  paid  for  it. 

The  difference  between  the  cost  and  the  face  of  the  note 
is  the  discount;  it  is  more  or  less,  according  to  the  time 
the  note  has  to  run. 

REVIEW. — 317.  If  a  note  docs  not  contain  these  words,  and  is  not 
paid  when  due,  what  is  the  consequence?  What  is  the  day  of  maturity  T 
<18.  If  a  note  is  payable  "  on  demand,"  when  is  it  legally  due? 


DISCOUNTING   NOTES. 


250 


The  time  to  run  is  the  number  of  days  from  the  day  the 
note  is  discounted,  to  the  day  the  note  is  legally  due, 
counting  the  latter,  but  not  the  former. 

REMARK. — The  time  to  run  is  taken  in  days  by  banks,  because  it 
IB  to  their  advantage;  if  it  were  taken  in  months  and  days,  each 
month  would  have  to  be  considered  30  days  in  calculating  the  dis- 
count, whereby  a  day  would  be  lost  in  each  month  of  31  days. 

ART.  320.  In  determining  the  day  of  maturity  and  tht 
lime  to  run,  it  is  convenient  to  use  this 

TABLE 
SHOWING  THE  NUMBER  OP  DAYS  PROM  ANY  DAY  OP 


CH 

% 

5- 

5 
P 
T< 

> 

*o 

;i 

S 
3 

2l5 
276 

CH 

3 
?• 

2T5 
245 

«H 

e^ 
«T 

1st 

215 

> 

s 

°P 

t» 

a> 

-o 

122 
153 

p 
r- 

&5 

o 
<J 

"61 

92 

O 

o 
p 

To  the 
same  day 
of  next 

365 
31 

334 

365 

306 
337 

275 
306 

153 
184 

92 
123 

31 
"62 

Jan. 

Feb. 

59 

28 

365 

334 

304 

273 

243 

212 

181 

151 

120 

90 

Mar. 

90 

59 

31 

365 

335 

304 

274 

243 

212 

182 

151 

121 

April.  ; 

120 

89 

61 

SO 

365 

334 

304 

273 

242 

212 

181 

151 

May. 

151 

120 

92 
122 
153 

61 
91 

122 

31 
61 
92 

365 
30 
61 

335 
365 
31 
62 
92 

304 
334 
365 

273 

243 
273 
304 

212 
242 
273 

182 
212 
243 
274 
304 

June. 

181 
212 
243 
373 

150 
181 

303 
334 

July. 

Aug. 

212 
242 

184 
214 

153 
183 

123 
153 

92 
122 

31 
61 

365 

335 
365 

304 
334 

Sept, 

30 

Oct. 

304 

273 

245 

275 

214 

184 

153 

123 

92 

61 

31 

365 

335 

Nov. 

334 

303 

244 

214 

183 

153 

122 

91 

61 

30 

365 

Dec. 

REMAKE. — In  leap  years,  if  the  last  day  of  February  is  included 
ja  the  time,  1  day  must  be  added  to  the  number  obtained  from  the 
table. 

A  note  maturing  Sept.  13,  is  discounted  June  24  pre 
vious :  what  is  the  time  to  run  ? 

SOLUTION. — From  June  24  to  Sept.  24,  by  the  table,  is  92  da.;  the 
13th  being  11  days  before  the  24th,  will  give  92  — 11,  =  81  da.  AM 
for  the  time  to  run. 

REVIEW. — 318.  If  it  is  payable  a  certain  time  after  date,  what  is  th» 
rule  for  finding  when  it  is  legally  duo?  Which  day  is  not  counted? 
What  are  days  of  grace  ?  Why  so  called  ?  When  is  a  note  nominally 
due  ?  When  legally  due  ?  What  are  the  months  ?  What  is  said  of  note* 
drawn  for  months  ?  How  is  this  irregularity  avoided  in  short  notes? 


260  RAY'S   HIGHER   ARITHMETIC. 


ART.  321.  CASE  I.  To  find  the  discount  and  proceeds 
of  a  Note  of  short  date,  custom  has  sanctioned  this 

RULE. 

Take  the  face  of  the  note  as  a  principal,  the  rate  of  discount 
as  the  rate  of  interest,  and  calculate  the  interest  for  the  time  Hie 
note  has  to  run ;  tliis  will  be  the  discount  of  the  note,  and  if  it 
is  subtracted  from  the  face  of  the  note,  the  remainder  will  *>< 
the  proceeds  or  cost  of  the  note. 

NOTES. — 1.  The  discount  on  short  notes,  resembles  the  discouui 
on  money,  stocks,  bonds,  &c.,  being  calculated  on  the  nominal  value 
or  face,  but  differs  from  the  discounts  on  debts  and  long  notes, 
(Art.  312) :  the  former  is  called  Bank  discount;  latter,  true  discount. 

2.  Since  the  face  of  every  note  is  a  debt  due  at  a  future  time, 
its  cost  ought  to  be  the  present  worth  of  that  debt,  and  the  bank 
discount  to  be  the  same  as  the  true  discount. 

As  it  is,  the  former  is  greater  than  the  latter ;  /or,  bank  discount  is 
the  interest  on  the  face  of  the  note,  while  true  discount  is  the  interest  on 
the  present  worth,  which  is  always  less  than  the  face.  Hence,  their 
difference  is  the  interest  of  the  difference  between  the  Present  Worth 
tad  face;  that  is,  the  interest  of  the  true  discount. 

3.  In  calculating  the  discount  on  notes,  use  generally  the  2d  rule 
(Art.  307).     The  operation  may  often  be  shortened,  by  recollecting 
that  when  the  two  right  hand  figures  of  dollars  are  pointed  off,  the 
result  is  the  interest  of  the  principal,  for  60  da.  at  6  ft,    for  72  da. 
at  6  %,    for  45  da.  at  8  ft,    for  35  da.  at  10  ft,    or  for  any  other 
number  of  days  and  rate,  whoso  product  is  3(30;  so  that  the  interest 
at  any  rate,  can  be  frequently  got  in  this  way  by  aliquot  parts, 
without  first  getting  it  at  G  ft. 

Find  the  day  of  maturity,  the  time  to  run,  and  the  pro- 
ceeds of  the  following  notes. 

1.  $792/05  CINCINNATI,  Jan.  3d,  1854. 

Six  months  after  date,  I  promise  to  pay  to  the  order  of 
Willis  &  Markham,  seven  hundred  ninety-two  fVo  dollars,  at 
the  Commercial  Bank,  value  received.  H.  WIIITTAKER. 

Discounted,  Feb.  18,  at  G  ft. 
Ans.  Due,  July  3!6;  138  da.  to  run;  Pro.  $774.27 


RKVIBW. — 319.  What  is  discounting  notes?  By  whom  is  it  done? 
What  kind  of  notes  are  discounted  ?  What  is  the  proceeds  or  cost  of  a 
note?  Tho  discount?  What  is  the  timo  to  run?  320.  Show  the  uso  of 
the  table.  What  is  said  of  leap  years?  321.  What  is  the  rule  for  die 
counting  short  notes?  What  does  bank  discount  resemble 7 


DISCOUNTING  NOTES.  261 

REMARK. — The  date  before  the  line,  in  July  3|6J  is  when  th« 
note  is  nominally  due,  the  other  when  it  is  legally  due. 

2.  $1066Tyo  LOUISVILLE,  May  19,  1855. 
Value  received,  ninety  days  after  date,  I  promise  to  pay 

Thomas  Beatty,  or  order,  one  thousand  sixty-six  iVo  dollars, 
•t  the  City  Bank.v  G.  TV.  ALEXANDER. 

Discounted  June  8,  at  6  %. 

Ans.  Due,  Aug.  17|2o;  73 da.  to  run;  Pro.  $1053.77 

3.  $1962  r«f5  NEW  YORK,  July  26,  1850. 
Value  received,  four  months  after  date,  I  promise  to  pay 

B   Thoms,  or  order,  one  thousand  nine  hundred  sixty-two 
rVff  dollars,  at  the  Chemical  Bank.  E.  WILLIAMS. 

Discounted  Aug.  26,  at  1%. 

Ans.  Due,  Nov.  26|29;  95da.  torun;  Pro.  $1926.70 

4.  $543A85  CINCINNATI,  Oct.  30,  1855. 
Thirty  days  after  date,  I  promise  to  pay  to  the  order  of 

Baker  &  Goodal,  five  hundred  forty-three  -,-Vb  dollars,  value 
received.  T.  H.  SHORT. 

Discounted  Oct.  31,  at  1  %  a  month. 

Ans.  Due,  Nov.  29 1  Dec.  2;  32da.  torun.    Pro.$537.88 

5.  $2672TVo  PHILADELPHIA,  March  10,  1852. 
Nino  months  after  date,  for  value  received,  I  promise  to 

pay  Edward  H.  King,  or  order,  two  thousand  six  hundred 
seventy-two  iVs  dollars.  JEREMIAH  BARTON. 

Discounted  July  19,  at  6  pe> 
Ans.  Due,  Dec. » °| ,  3  ;  147 da.  to  run.     Pro.  $2606.71 

6.  $804T3o9fl  COLUMBUS,  Aug.  12,  1854. 
Three  months   after  date,  I  promise  to  pay  at  the  City 

Bank  of  Columbus,  eight  hundred  four  iVo  dollars  to  the 
order  of  Irwin  &  Lee,  value  received.     JOSIAH  NEEDHAM. 
Discounted  September  3,  at  6  %. 
Ans.  Due,  Nov.  12|16  j  73  da.  to  run.    Pro.  $794.60 

REVIEW. — 321.  What  kind  of  notes  are  discounted  by  this  rule?  How 
ere  long  notes  running  for  one  or  more  years  discounted?  What  ought 
the  cost  of  a  note  to  bo?  What  would  the  bank  discount  then  be?  Which 
is  greater,  bank  discount  or  true  discount?  Why?  How  much  ?  What 
rule  of  interest  is  generally  used  in  notes?  How  may  the  operation  often 
be  shortened  ? 


262  RAYS  HIGHER  ARITHMETIC. 

7.  $3886.  ST.  Louis,  Jan.  31,  1853. 
One  month  after  date,  wo  jointly  and  severally  promise 

to  pay  C.  McKnight,  or  order,  three  thousand  eight  hundred 
eighty-six  dollars,  value  received.  T.  MONEOE, 

Discounted  Jan.  31,  at  U  %  a  month.  I.  FOSTER. 

Am.  Due,  Feb.  28  [Mar.  3;  31da.  to  run.  Pro.  $3825.77 

REMARK.  — A  note,  drawn  by  two  or  more  persons,  "jointly  and 
•everally,"  may  be  collected  of  either  of  the  makers,  but  if  the  words 
"jointly  and  severally  "  are  not  used,  it  can  only  be  collected  of  tht 
makers  as  a  firm  or  company. 

8.  $1425.  NASHVILLE,  April  11,  1853. 
For  value  received,  eight  months  after  date,  we  promise 

to  pay  Henry  Hopper,  or  order,  one  thousand  four  hundred 
twenty-five  dollars,  with  interest  from  date  at  6  per  cent, 
per  annum.  NIXON  &  MARSH. 

Discounted  June  15,  at  6  f0- 

Ans.  Due,  Dec.  »»|14;  182  da.  to  run.   Pro.  $1437.73 

N.  B. — Observe  that  the  face  of  this  note  is  the  amount  of  $1425 
for  8  mon.  3  da.  at .  G  fc. 

9.  $3703T8c4o  BALTIMORE,  June  6,  1850. 
For  value  received,  four  months  after  date,  we  promise 

to  pay  to  the  order  of  Jones  &  Newcome,  three  thousand 
seven  hundred  and  three  dollars  and  eighty-four  cents,  at 
the  Savings  Bank.  THOMAS  SHARPE  &  Co. 

Discounted  June  18,  1850,  at  1  %  a  month. 
Ans.  Due,  Oct.  «|9 ;  113da.  to  run.     Pro.  $3564.33 

10.  $813T6o°o  DAYTON,  May  31,  1856. 
For  value  received,  sixty  days  after  date,  I  promise  to 

pay  to  the  order  of  Hiram  Wells  &  Co.,  eight  hundred 
thirteen  dollars  sixty  cents.  JAMES  T.  FISHER. 

Discounted  May  31,  1856,  at  2  fc  a  month. 

Am.  Due,  July  30 1  Aug.  2;  63 da.  to  run.    Pro.  $779.43 

11.  $737T4o°o-        '  BOSTON,  Feb.  14,  1856. 
Value  received,  two  months  after  date,  I  promise  to  paj 

to  J.  K.  Eaton,  or  order,  seven  hundred  thirty-seven  7*0  / 
dollars,  at  the  Suffolk  Bank.  WILLIAM  ALLEN. 

Discounted  Feb.  23,  at  \§%. 
Ans.  Due,  April  14|I7;   54 da.  to  run.     Pro.  $726.34 


DISCOUNTING  NOTES.  263 


12.  §4085T3o0o-  NEW  ORLEANS,  Nov.  £0,  1855. 
Value  received,  six  months  after  date,  I  promise  to  pa 

John  A.  Westcott,  or  order,  four  thousand  eighty-five  T? 
dollars,  at  the  Planters'  Bank.  E.  WATERMAN. 

Discounted  Dec.  31,  1855,  at  5  %. 
Ana.  Due  May  ^(^ 3, 1856;  144da.  to  run.    Pro.$4003.50 

13.  $2623TVo-  CINCINNATI,  Aug.  7,  1854. 
For  value  received,  eighteen  months  after  date,  I  promise 

to  pay  to  the  order  of  Jonathan  Evans,  two  thousand  six 
hundred  twenty-three  T5^  dollars,  with  interest  at  ten  per 
cent,  per  annum.  MORRIS  TALBOT. 

Discounted  June  24,  1855,  at  1^  %  a  month. 
.4ns.  Due,  Feb.  7|10,1856;  231da.torun.    Pro.$2728.61 

ART.  322.  It  may  be  inquired,  what  rate  of  interest  is 
paid,  when  a  note  is  discounted.  The  proceeds  of  the  note 
is  the  sum  received  or  borrowed,  and  is,  therefore,  the  prin- 
cipal, while  the  discount  is  its  interest  for  the  time  the  note 
runs  ;  so  that  having  the  principal,  time,  and  interest,  the 
rate  of  interest  can  be  found  as  in  Art.  309. 

It  is  simpler,  however,  to  leave  the  face  of  the  note  out 
of  view,  and  proceed  thus  : 

TO    FIND    THE   RATE    OF    INTEREST  WHEN   A   NOTE   18 
DISCOUNTED. 

RULE. — Assume  $100  for  the  face  of  the  note,  and  on  this  sup- 
position determine  the  discount  and  proceeds  for  the  time  it  has  to 
run ;  the  former  will  be  the  interest ;  the  latter,  the  principal;  and 
the  lime  to  run,  the  time :  from  which  the  rate  of  interest  can  be 
found  by  Case  II  of  Simple  Interest. 

What  is  the  rate  oi  interest,  when  a  sixty  day  note  is 
discounted  at  2  %  a  month  ? 

SOLUTION. — For  every  $100  in  the  face  of  the  note,  the  discount 
for  03  da.  at  2  %  a  month,  is  $4.20,  and  the  proceeds  $95.80;  then 
$95.80  being  the  principal,  and  $4.20  its  interest  for  63  da.,  the  rate 
of  interest  is  found  by  Art.  309  to  be  25$%%,  per  annum. 

REVIEW. — 322.  When  a  note  is  discounted,  what  is  the  principal  or 
earn  borrowed?  What  its  interest  ?  How  may  the  rate  of  interest  then  be 
foun.l  ?  What  tnav  be  loft  out  of  view  in  this  calculation  ? 


264  RAY'S   HIGHER   ARITHMETIC. 


WHAT  IS  THE    HATE  01'  INTEREST, 

1.  "When  a  30  da.  note  is  discounted  at  1,  1|,  lA,  2  %  a 
mon.?  Ans.  12$|g,  lo^es,  ISiVaV  24i8883  %  per  annum. 

2  When  a  60  da.  note  is  discounted  at  6,  8,  10%  per 
annum?  Ans.  6nm>,  S-iVs,  10  5%  %  per  annum. 

8.  When  a  90  da.  note  io  discounted  at  2,  2A,  3  %  a  mon.? 
^4?is.  25llf,  32rV;i,  39g5l  %  per  annum. 

4.   When  a  note  running  1  yr.  is  discounted  at  5,  6.  Y,  S, 

9,10,12%?  ' 


RE  u  ARK.  —  It  may  seem  unnecessary  to  regard  the  time  the  note 
has  to  run,  in  determining  the  rate  of  interest;  but,  a  comparison  of 
examples  1  and  3,  shows  that  a  90  da.  note,  discounted  at  2  fe  a 
month,  yields  a  higher  rate  of  interest  then  a  30  da.  note  of  the  same 
face,  discounted  at  '2  %  a  month.  The  discount,  at  the  same  rate,  on 
all  notes  cf  the  same  face,  varies  as  the  time  to  run,  and  if  in  each 
case,  it  was  referred  to  the  same  principal,  the  rate  of  interest  would 
oe  the  same,  but  when  the  discount  becomes  tarter,  me  proceeds  or 
principal  to  which  it  is  referred,  becomes  smaller,  and  therefore  the 
rate  of  interest  corresponding  to  any  rate  of  discount  increases  with  the 
time  the  note  has  to  run.  Hence,  the  profit  of  the  discounter  is  greater 
proportionally  on  long  notes  than  short  ones,  at  the  same  rate. 

ART.  323.  Discounting  Notes  is  an  application  of  Simple 
[nterest  :  the  face  of  the  note  is  the  principal  ;  the  discount 
is  the  interest  ;  the  rate  of  discount  is  the  rate  of  interest;  and 
the  time  to  run  is  the  time. 

Hence,  it  may  be  divided  into  cases  corresponding  to, 
and  solved  like  those  of  simple  interest.  The  only  case  of 
sufficient  importance  to  require  distinct  notice,  is 

CASE  II. 

Given,  the  proceeds,  time,  and  rate  of  discount,  to  find 
the  face  of  the  note. 

RULE.  —  Assume  $1  for  the  face  of  the  note  ;  determine  the  pro- 
ceeds on  this  supposition,  and  divide  the  given  proceeds  by  it. 

REVIEW.  —  322.  Give  the  rule.  Analyze  tho  example.  Can  the  time 
to  run  be  left  out  of  view  in  determining  tho  rate  of  interest  corresponding 
to  any  rate  of  discount?  Why  not?  What  does  the  rate  of  interest  in- 
crease with?  What  sort  of  notes  arc  most  profitable  to  the  discounter, 
then,  provided  tho  rates  of  discount  are  the  same?  323.  What  is  Case  27 
The  Rule?  The  proof  ?  Solve  the  example. 


DISCOUNTING   NOTES.  2OH 


PROOF. — Discount  the  note  thus  found ;  if  it  yields  the 
given  proceeds,  the  work  is  right. 

For  what  sum  must  a  60  da.  note  be  drawn,  to  yield 
£1000,  when  discounted,  at  6  %  per  annum? 

SOLUTION. —  For  every  $1  in  the  face  of  the  note,  the  proceeds, 
ty  Case  I,  (Art.  321),  is  $.989.3 ;  hence,  there  must  be  as  many  dol- 
lars in  the  face  as  this  sum,  §.989-3,  is  contained  times  in  the  given 
proceeds,  §1000:  this  gives  §1010.01  for  the  face  of  the  note. 

1.  Find  the  face  of  a  30  da.  note,  which  yields  $1G60 
when  discounted  at  lj  c/0  a  mon.  Ans.  §1077.08 

2.  The  face  of  a  GO  da.  note,  which,  discounted  at  6  % 
per  annum,  will  yield  $800.  Ans.  $808.49 

3.  The  face  of  a  90  da.  note,  which,  discounted  at  7  % 
per  annum,  yields  $1235.40.  Ans.  $1258.15 

4.  The  face  of  a  4  mon.  note,  which,  discounted  at  1  % 
a  month,  yields  $3375.  Ans.  $3519.29 

5.  The  face  of  a  6  mon.  note,  which,  discounted  at  10  % 
per  annum,  yields  $4850.  Ans.  $5109.75 

G.  The  face  of  a  60  da.  note,  discounted  at  2%  a  month, 
to  pay  $768.25.  Ans.  $801.93 

7.  The  face  of  a  40 da.  note,  which,  discounted  at  8%, 
yields  $2072.60.  Ans.  $2092.60 

8.  The  face  of  a  30  da.  and  90  da.  note,  to  net  $1000 
when  discounted  at  6  %• 

AM.  $1005.53  at  30  da.;  $1015.74  at  90  da. 

ART.  324.  To  find  the  rate  of  discount  corresponding 
to  a  given  rate  of  interest. 

RULE. — Assume  $100yb?-  the  2>rocecds  ;  find  its  interest  and 
amount  at  the  gicen  rate,  for  the  lime  the  note  runs ;  the  latter  is 
the  face  of  the  note,  and  the  former  is  the  discount,  or  interest  on 
that  face  for  the  time  the  note  runs ;  the  rale  of  interest  will  then 
be  found  by  Case  II  of  Simple  Interest,  and  will  be  the  rate  of 
discount  required. 

If  I  wish  to  get  interest  at  the  rate  of  20  %  per  annum, 
it  vdiat  rate  should  I  discount  60  da.  notes? 

SOLUTION. — For  every  §100  in  the  proceeds,  I  wish  to  get  in- 
terest at  20  r/0  per  annum,  which  for  03  days  is  §3.50,  interest,  and 


R  E  v  i  B  w. — 324.  What  is  the  rule  for  finding  the  rato  of  discount  cor- 
responding to  a  given  rate  of  interest?     Analyze  the  example. 

23 


RAPS   HIGHER   ARITHMETIC. 


$103.50,  tmount  take  $103.50  as  the  face  of  the  note,  or  principal, 
and  §3.50  as  »ho  discount,  or  interest  of  that  principal  for  U3  d;iys, 
aud  find  the  rate  by  Case  II  of  Simple  Interest, 


WHAT    RATES    OF    DISCOUNT 

1.  OD  30  da.  notes,  yield  10,15,  20,  30,  40,  50  %  inter- 

,o  Oll01-f/ir>4<i-in39I       OQ  27        o  o  1  8  i     *  7  2  Q  3  -,/ 

St?       Ans.  IhvjTT,  14sii,  19817,   ^9T3'7>  oO3TT,  4  (  25i  %. 

2.  On  60  da.  notes,  yield  6,  8,  10,  12,  18,  24  %  interest? 


Ans.  5Ag|f,  7iU,  9| 


g?, 


3.  On  90  da.  notes,  yield  1,  1^,2,2.4,  3,4^  amon.int.? 

i          -l-i     659      -I  >7    4  1  9      Q  a  I  0  ft     o  7  :*  *>  3     o  o  '  0  -  4      ,f  o  1  'J  8  ,,/• 

.4ns.  lljiiaT,  I/joss,  ^-rt?,  -  <  iai,  oliiijgf,  4Jas  i  %• 

4.  On  notes  running  1  yr.  without  grace,  yield  5,  6,  7, 
8,  9,  10,  12,  15,  18,  20,  25  %  interest? 

A«*.  4H,  5ff,  6T5c?T,  T^,  8T%,  9iT,  101,  13 
16§,  20%. 


PARTIAL   PAYMENTS. 

AJIT.  325.  A  note,  not  paid  at  maturity,  draws  interest 
at  the  legal  rate,  from  the  day  it  is  due  till  it  is  paid  ; 
but  if  it  contains  the  words  "  with  interest,"  it  draws  in- 
terest from  date.  Any  sums  paid  on  account  of  the  note, 
arc  indorsed  with  their  respective  dates,  and  are  called 
Partial  Payments.  In  such  cases,  the  balance  due  on 
settlement  is  foiv.n}  by  the 

U.  S.  RULE  FOR  PARTIAL  PAYMENTS. 

•'Apply  the  paymint,  in  the  first  place,  to  the  discharge  of 
the  interest  t  ten  dm:;  if  the  payment  EXCEEDS  the  ind'sexl,  the 
surplus  r/oef  toward  discharging  the  principal,  and  the  subse- 
quent inter, st  is  to  be  computed  on  the  balance  of  principal 
remaining  due. 

"If  t1^^  paymeTl  be  LESS  than  the  interest,  the  surplus  of 
interest  must  not  le  taken  to  augment  the  principal;  but  interest 
continvis  on  the  firmer  principal,  until  the  period  when  the  pay- 
ments taken  together  exceed  tlie  interest  due,  and  then  the  surplus 


REVIEW. — 325.  If  a  note  is  not  paid  at  maturity,  what  is  tho  conse- 
quence? If  it  co' itains  the  words  "with  interest,"  when  does  interest 
commence?  What  are  partial  payments?  What  is  the  United  States  rule 
for  casting  interesl  on  notes  and  bonds  when  partial  payments  have  hoen 
mado  ?  On  what  principle  \t  this  rule  founded  ? 


PARTIAL  PAYMENTS.  207 

is  to  be  applied  toward  discharging  the  principal,  and  interest  u 
to  be  computed  on  the  balance  as  aforesaid" 

I!  KM  AUK. — 1.  This  rule,  is  adopted  by  the  Courts  of  the  U.  3. 
and  of  most  of  the  States  ;  it  is  on  the  principle  that  neither  interetl 
nor  payment  shall  draw  interest. 

2.  It  is  worthy  of  remark,  that  the  whole  aim  and  tenor  of  legis- 
lative enactments  and  judicial  decisions  on  questions  of  interest, 
have  been  to  favor  the  debtor,  by  disallowing  compound  interest,  and 
yet  this  very  rule  fails  to  secure  the  end  in  view,  and  really  maintaini 
and  enforces  the  principle  of  compound  interest  in  a  most  objcctkn- 
able  shape ;  for  it  makes  interest  due,  (not  every  year  as  compound 
interest  ordinarily  does)  but  as  often  as  a  payment  is  made;  by  which 
it  happens  that  the  closer  the  payments  are  together,  the  greater  the  loss 
of  the  debtor,  who  thus  suffers  a  penalty  for  his  very  promptness. 

To  illustrate,  suppose  the  note  to  be  for  $2000,  drawing  interest  at 
6  fe,  and  the  debtor  pays  every  month  §10,  which  just  meets  the 
interest  then  due  ;  at  the  end  of  the  year  he  would  still  owe  $2000. 
But  if  he  had  invested  the  $10  each  mouth,  at  G  fa,  he  would  have 
had,  at  the  end  of  the  year,  $123.30  available  for  payment,  while  the 
debt  wou'd  have  increased  only  $120,  being  a  difference  of  $3.30  in 
his  favor,  and  leaving  his  debt  $1996.70,  instead  of  $2000. 

3.  To  find  the  difference  of  time  between  two  dates  on  the  note, 
reckon  by  years  and  months  as  far  as  possible,  and  then  count  the 
days ;  as  in  the  rule  Art.  318. 

$850.  CINCINNATI,  April  29,  1850. 

Ninety  days  after  date,  I  promise  to  pay  Stephen  Ludlow, 
or  order,  eight  hundred  fifty  dollars,  with  interest;  value 
received.  CHARLES  K.  TAYLOR. 

Indorsements.— Oct.  13,  1850,  $40;  June  9,  1851,  $32;  Aug.  21, 
1851,  $125;  Dec.  1,  1851,  $10;  March  16,  1852,  $80. 

What  was  due  Nov.  11,  1852? 

SOLUTION. — Interest  on  face  ($850)  from  April  29  to 

Oct.  13,  1850,  being  5  mon.  14  da.,  at  6  %  per  annum,  $23.238 

850. 
tfhole  sum  due  Oct.  13,  1850,    .         .         .....    873.233 

Payment  to  be  deducted,  ....         40. 

Balance  due  Oct.  13,  1850 833.233 

REVIKW. — 225.  How  docs  the  rule  allow  compound  interest  ?  Illn»- 
trate  by  an  example.  IIow  find  the  difference  of  time  between  two  dates, 
nn  the  note  T 


208  RAY'S   HIGHER   ARITHMETIC. 

Interest  on  balance  ($833.233)  from  Oct.  13,  1850,  to 

June  9,  1851,  being  7  mon.  27  da., 32.913 

Payment  not  enough  to  meet  the  interest, 32. 

Surplus  interest  not  paid  June  9,  1851, .913 

Interest  on  former  principal  ($833?233)  from  June  9, 

1851,  to  Aug.  21,  1851,  being  2  mon.  12  da.,      .     .     .        9.999 

Whole  interest  due  Aug.  21,  1851, 10.912 

833.233 

Whole  sum  due  Aug.  21,  1851, 844.145 

Payment  to  be  deducted, 125. 

Balance  due  Aug.  21,  1851, 719.145 

Interest  on  the  above  balance  ($719.145)  from  Aug.  21, 

1851,  to  Dec.  1,  1851,  being  3  mon.  10  da.,    ....  11.986 

Payment  not  enough  to  meet  the  interest, 10. 

Surplus  interest  not  paid  Dec.  1,  1851, 1.986 

Interest  on  former  principal  ($719.145)  from  Dec.  1, 

1851,  to  March  1C,  1852,  being  3  mon.  15  da.,    .     .     .  12.585 

Whole  interest  due  March  16,  1852 14.571 

719.145 

Whole  sum  due  March  16,  1852,     ........     733.716 

Payment  to  be  deducted, 80. 

Balance  due  March  16,  1852, 653.716 

Interest  on  Balance  from  March  16,  1852,  to  Nov.  11, 

1852,  being  7  mon.  26  da., 25.713 

Balance  due  on  settlement,  Nov.  11,  1852, $679.43 

1.  $304TYo-  CHICAGO,  March  10,  1852. 
For  value  received,  six  months  after  date,  I  promise  to 

pay  G.  Riley,  or  order,  three  hundred  and  four  yVo"  dollars. 
No  payments.  IL  McM.VKIN. 

What  was  due  Nov.  3,  1853?  Am.  $325.63 

2.  $2250.  LOUISVILLE,  Dec.  G,  1850. 
For  value  received,  one  year  after  date,  I  promise  to  pay 

/Vlbert  Rogers,  or  order,  two   thousand  two  hundred  and 
fifty  dollars,  with  interest.  NATHAN  PniLirs. 

Indorsed:  April  13,  18-52,  $1000. 

What  was  due  Aug.  19,  1854?  Ans.  $1634.63 

3.  $429fo°o-  INDIANAPOLIS,  April  13,  1853. 
On  demand,  I  promise  to  pay  W.  Morgan,  or  order,  four 

hundred   and   twenty-nine  iVo"  dollars,  value  received. 

R.  WILSON. 


PARTIAL  PAYMENTS.  269 


Indorsed:  Oct.  2, 1853,  $10;  Dec.  8, 1853,  $60;  July  17, 1854,  §200. 
What  was  due  Jan.  1,  1855?  Am.  $195.06 

4.  $1750.  NEW  YORK,  Nov.  22,  1852. 
For  value  received,  two  years  after  date,  I  promise  to  pay 

to  the  order  of  Spencer  &  Ward,  seventeen  hundred  and  fifty 
dollars,  with  interest  at  7  per  cent.  JACOB  AViNSTON. 

Indorsed:  Nov.  25,  1854,  $500;  July  18,  1855,  $50;  Sept.  1,  1864 
$600;  Dec.  28,  1855,  $75. 

What  was  due  Feb.  10,  1856?  Ans.  $879.71 

5.  $4643T5o°o  DAYTON,  Ohio,  March  7,  1851. 
For  value  received,  three  years  after  date,  I  promise  to  pay 

R.  Banks,  or  order,  forty-six  hundred  and  forty  three  -ffa 
dollars,  with  interest  at  10  per  cent.  W.  G.  BROOKS. 

Indorsed:  June  25,  1854,  $1000;  Nov.  1,  1854,  $500;  Jan.  12,  1855, 
$2500;  Sept.  4,  1855,  $1350;  May  10,  1856,  $150. 

What  was  due  July  1,  1856?  Ans.  $1189.86 

6.  $540.  BALTIMORE,  Jan.  11,  1849. 
Eighteen  months   after   date,  wo  promise  to  pay   Silas 

Greene,  or  order,  five  hundred  and  forty  dollars,  with  in- 
terest, value  received.  EVANS  &  HART. 

Indorsed:  Nov.  23,  1850,  $125;  March  5,  1851,  $35;  Feb.  27,  1852, 
$25;  May  31,  1852,  $80;  Oct.  16,  1852,  $100. 

What  was  due  March  1,  1853  ?  Ans.  $291.60 

7.  $2500.  PHILADELPHIA,  Aug.  8,  1850. 
For  value  received,  nine  months  after  date,  I  promise  to 

pay  Abijah  Warren,  or  order,  two  thousand  five  hundred 
dollars,  with  interest  at  6  per  cent.  HENRY  CROSS. 

Indorsed:  Nov.  1,  1851,  $400;  Dec.  14,  1852,  $100;  July  6,  1853, 
$50;  Oct.  21,  1854,  $750;  April  18,  1855,  $500. 

What  was  due  July  1,  1855?  Ans.  $1362.04 

8.  $6875.  BOSTON,  Oct.  22,  1852. 
For  value  received,  four  months  after  date,  we  promise  to 

pay  Augustus  King,  or  order,  six  thousand  eight  hundred 
and  seventy-five  dollars.  DAVIS  &  UNDERWOOD. 

Indorsed:  Feb.  25,  1853,  $2000;  Jan.  1,  1854,  $245;  April  1,  1854, 
$76;  July  10,  18-31,  $1500;  Dec.  '26,  1854,  $95;  May  12,  1855,  $1200 
Bept.  29/1855,  $50. 

What  was  due  Jan.  1,  1856?  Ans.  $2376.64 


RAY'S   HIGHER   ARITHMETIC. 


0.  $550.  ST.  Louis,  June  17,  1848. 
For  value  received,  one  year  after  date,  I  promise  to  pay 

to  the  order  of  Ross  &  Wade,  five  hundred  and  fifty  dollars, 
with  interest.  TIMOTHY  GORDON. 

Indorsed:  Aug.  19,  1848,  $100;  Dec.  1,  1848,  $40;  Feb.  8,  1850, 
$30;  Sept.  27,  1850,  $200;  Jan.  4,  1852,  $10;  Mar.  1,  1852,  $60. 

What  was  due  Aug.  28,  1852  ?  Am.  $195.87 

CONNECTICUT   RULE. 

ART.  326.  "  Compute  the  interest  to  the  time  of  the  first 
payment,  if  that  lie  one  year  or  more  from  the  time  that  interest 
commenced;  add  it  to  the  principal,  and  deduct  the  payment 
from  the  sum  total. 

"  If  there  be  after  payments  made,  compute  the  interest  on  the 
balance  due  to  the  next  payment,  and  then  deduct  the  payment 
as  above ;  and  in  like  manner  from  one  payment  to  another,  till 
all  the  payments  are  absorbed:  Provided,  the  time  between  out 
payment  and  another  be  one  year  or  more. 

"  But  if  any  payment  be  made  before  one  year's  interest  hath 
accrued,  then  compute  the  interest  on  the  ^n'ncjpcZ  sum  due  on 
the  obligation,  for  one  year,  add  it  to  the  principal,  and  compute 
the  interest  on  the  sum  paid,  from  the  time  it  was  paid,  up  to  the 
end  of  the  year ;  add  it  to  the  sum  paid,  and  deduct  that  sum 
from  the  principal  and  interest  added  as  above.  ( See  Note. ) 

"  If  any  payment  be  made  of  a  less  sum  than  the  interest 
arisen  at  the  time  of  such  payment,  no  interest  is  to  be  com- 
puted, but  only  on  the  principal  sum,  for  any  period." 

NOTE. — "If  a  year  does  not  extend  beyond  the  time  of  payment ;  but 
if  it  does,  then  find  the  amount  of  the  principal  remaining  unj>aid  up  to  the 
time  of  settlement,  likewise  the  amount  of  the  payment  or  payments  from 
the  time  they  were  paid  to  the  time  of  settlement,  and  deduct  the  sum  of 
these  several  amounts  from  the  amount  of  the  principal. 

1.  What  is  due  on  the  2d,  3d,  and  4th  of  the  preceding 
notes,  by  the  Connecticut  rule? 

4n*.  2d,  $1634.63;   3d,  $194.54;  4th,  $877.95 

VERMONT   RULE. 

ART.  327.  Find  the  simple  interest  of  the  principal  from 
the  time  it  begins  to  draw  interest  to  the  time  of  settlement,  and 
add  it  to  the  principal.  Do  the  same  for  each  payment.  .Add 
together  the  amounts  of  the  several  payments,  and  deduct  the  sum 
from  the  amount  of  the  principal.  The  remainder  will  be  the 
balance  due  on  settlement. 


EXCHANGE.  271 


NOTE. — This  rule  is  of  frequent  use,  when  settlement  takes  place 
a  year  or  less  from  the  time  interest  begins. 

1.  What  is  due  on  the  2d,  3d,  and  4th  of  the  preceding 
notes,  by  the  Vermont  rule? 

Am.  2d,  $1608.88;  3d,  $193.50;  4th,  $855.79 

REMARK. — The  following  has  been  recommended  as  a  more  equi- 
table RDLE  for  Partial  Payments,  than  any  in  use. 

"  Starting  at  the  time  interest  begins,  find  the  Present  Worth  by  Simple 
tnterest  of  each  payment ;  deduct  the  sum  of  these  Present  Worths  from 
the  Principal :  the  amount  of  the  balance,  by  Simple  Interest,  to  the  day 
of  tettlement,  will  be  the  turn  then  due." 

The  principle  of  this  rule  is,  each  payment  discharges  a  part  of  the 
principal  with  its  simple  interest  to  the  day  the  payment  is  made  ;  making 
interest  and  principal  due  at  the  same  time,  instead  of  the  interest 
all  due  first,  and  the  principal  afterward. 


XXVI.  EXCHANGE. 

ART.  328.  EXCHANGE  is  the  method  of  transmitting  money 
from  one  place  to  another  by  means  of  Bills  of  Exchange. 

If  the  places  are  in  different  countries  it  is  called  Foreign 
Exchange  ;  if  not,  Home,  Domestic  or  Inland  Exchange. 

Bills  of  Exchange,  also  called  drafts  or  checks,  are  writ- 
ten orders  for  the  payment  of  money. 

A  Sight  Bill,  is  one  payable  "at  sight." 

A  Time  Bill  is  payable  a  specified  time  after  sight,  or 
after  date. 

The  signer  of  the  bill,  is  the  maker  or  drawer. 

The  one  to  whom  the  draft  is  addressed,  and  who  is  re- 
quested to  pay  it,  is  the  drawee. 

The  one  to  whom  the  money  is  ordered  to  be  paid,  is 
the  payee. 

The  one  who  has  possession  of  the  draft,  is  called  the 
owner  or  holder ;  when  he  sells  it,  and  becomes  an  indorser, 
be  is  liable  for  payment.. 

RRVIEW. — 323.  What  is  exchange  ?  Foreign  exchange?  Home  ex- 
change  ?  What  arc  Bills  of  Exchange  ?  What  is  a  sight  bill  'I  A  time 
bill ?  Who  is  the  drawer  ?  The  drawee ?  The  payee?  Who  »n  tLa  holder .' 
What,  if  he  becomes  an  indorser? 


«>72  KAY'S   HIGHER   ARITHMETIC. 


A  special  indorsement  is  an  order  to  pay  the  draft  to  a  particular 
person  named,  who  is  called  the  indorsee,  as  "  Pay  to  F.  II.  Lee. — W. 
Harris/'  and  no  one  but  the  indorsee  can  collect  the  bill. 

When  the  indorsement  is  in  blank,  the  payee  merely  writes  his  name 
on  the  back,  and  any  one  who  has  lawful  possession  of  the  draft  can 
collect  it. 

If  the  drawee  promises  to  pay  a  draft  at  maturity,  he  writes  across 
the  face  the  word  "Accepted,"  with  the  date,  and  signs  his  name, 
thus:  "Accepted,  July  11,  1851. — II.  Morton."  The  acceptor  is  first 
responsible  for  payment,  and  the  draft  is  called  an  acceptance. 

ART.  329.  A  bill  of  exchange,  like  a  promissory  note,  may 
be  payable  "to  order,"  or  " bearer,"  and  is  subject  to  protest 
in  case  the  payment  or  acceptance  is  refused  :  it  is  also  en- 
titled to  the  3  days  grace,  whether  a  sight  or  time  bill. 

In  some  States,  drafts  are  not  entitled  to  days  of  grace. 

ART.  330.  The  following  are  forms  of  drafts  or  bills  of 
exchange. 

INLAND   DRAFT. 

$1500.  CINCINNATI,  Feb.  8,  1855. 

Please  pay,  at  sight,  to  Williams  &  Baker,  or  order,  fif- 
teen hundred  dollars,  value  received,  and  charge 

To  Kiyo  &  CLARK,  Brokers,  New  York.         THOMAS  ATKINS. 
FOREIGN   DRAFT. 

Exchange  £2000.  New  York,  May  21,  1853. 

At  sixty  days  sight  of  this  first  exchange  (second  and  third 
of  the  same  date  and  tenor  unpaid),  pay  to  the  order  of 
II.  Watts,  two  thousand  pounds,  without  further  advice. 

To  OEOROE  SPENCE,  Merchant,  Liverpool.      ELMER  &  JiATES. 

NOTES. — 1.  The  words  "value  received"  are  not  essential  to  a 
bill  of  exchange. 

2.  In  foreign  exchange,  three  separate  bills  are  generally  drawn, 
BO  that  if  one  or  two  are  lost,  the  other  may  reach  its  destination. 
When  one  bill  of  a  set  has  been  paid,  the  rest  are  void. 


KB  VIEW. — 328.  What  is  special  indorsement?  Give  an  example. 
What  is  the  consequence  of  a  special  indorsement  ?  What  is  an  indorse- 
ment in  blank?  Its  eonsequeuee?  M'hat  is  an  acceptance?  Give  un 
exiimpio.  What  is  the  effect  of  it?  329.  How  does  a  bill  of  exchange 
rcsi.-mt.lo  a  promissory  note?  AVli.it  is  said  of  protest?  Of  grace  ?  Of 
niglu-bilis  ?  ;;30.  Give  nn  example  of  an  inland  drafu  Of  u  foreign  dmft. 
What  is  said  of  the  words  "value  received"?  What  is  a  set  of  foreign 
ac^e  ?  When  one  of  a  set  is  paid,  what  is  the  consequence  ? 


EXCHANGE.  273 


ART.  331.  The  rate  of  exchange  is  a  rate  per  cent,  of  the 
face  of  the  draft. 

The  rate  of  exchange  between  two  places  depends  on  their  trade 
with  each  other ;  thus,  at  New  Orleans,  drafts  on  New  York  will  be 
at  a  premium,  more  or  less,  according  as  the  demand  at  New  Orleans, 
for  drafts  to  make  payments  in  New  York,  exceeds,  more  or  less,  the 
supply  furnished  by  parties  drawing  against  sums  due  them  in  New 
York. 

If  the  demand  is  less  than  the  supply,  exchange  on  New  York  is 
at  a  discount  in  New  Orleans,  more  or  less,  according  to  the  excess. 

If  the  demand  just  equals  the  supply,  exchange  is  at  par. 

ART.  332.  The  calculations  connected  with  inland  ex- 
change have  been  explained  in  Art.  276-279. 

The  computation  of  foreign  exchange  is  similar,  except 
that  it  is  necessary  to  express  the  money  of  one  country 
in  that  of  the  other. 

The  sum  mentioned  in  a  draft  on  a  foreign  country,  is  expressed 
in  the  money  of  that  country. 

In  calculating  the  premium,  discount,  or  cost  of  such  a  draft,  in  the 
home  currency,  it  is  necessary  not  only  to  have  tables  of  foreign 
money,  but  also  to  know  the  comparative  values  of  the  home  and 
foreign  currencies. 

The  _par  of  exchange  is  the  comparative  value  of  the  money 
of  two  countries,  depending  upon  the  amounts  of  pure  gold 
or  silver  they  contain  :  it?  must,  therefore,  remain  fixed,  as 
long  as  the  coins  retain  the  same  weight  and  fineness;  as, 
$4.86  (intrinsic  value,)  =  £1,  and  $1  =  5|  francs. 

The  course  of  exchange  is  the  par  of  exchange  after  allow- 
ing for  the  rate  of  exchange;  since  the  rate  of  exchange  if> 
variable  according  to  the  balance  of  trade,  the  course  of 
exchange  will  also  fluctuate  within  certain  limits,  being 
sometimes  above  and  sometimes  below  the  par,  as,  $1  = 
5.25  francs,  $4.87  (exchangeable  value,)  =  £1. 

REVIEW. — 331.  What  is  the  rate  of  exchange?  What  docs  it  depend 
on?  Example.  332.  To  what  subject  do  computations  in  inland  Exchange 
belong  ?  How  does  the  computation  of  foreign  exchange  differ  from  thnt 
of  inland  exchange  ?  How  is  the  face  of  a  foreign  draft  expressed  T 
What  must  be  known  to  find  iU  value  in  our  currency  T  What  is  the  paj 
of  exchange  between  two  countries  ?  What  does  it  depend  on  ?  Can  U 
•vet  change* 


274  RAY'S  HIGHER  ARITHMETIC. 


Since  the  course  of  exchange  allows  for  the  rate  of  ex- 
change, when  the  former  is  known,  the  cost  of  the  bill  ia 
found  by  reduction,  without  using  Percentage. 

If  the  rate  of  exchange  is  given,  a  reduction  of  curren- 
cies, and  a  calculation  in  Percentage,  are  both  required. 

NOTES. — 1.  For  Great  Britain  and  her  possessions,  the  RATE  ol 
exchange  is  given.  For  other  foreign  countries,  the  COURSE  of  ex- 
change is  given. 

2.  By  comparing  the  pure  metal  in  an  English  sovereign  with  that 
in  a  U.  S.  dollar,  it  is  found  that  20s.  or  £1  =  $4.86 ;  formerly  the 
silver  coin  of  the  U.  S.  was  purer  than  at  present,  and  then  20s.  or 
£1,  was  equal  to  $4|  or  $4.44|.  Instead  of  comparing  U.  S.  money 
with  sterling  money  at  the  rate  of  £1  =  $4.80,  it  is  customary  still 
to  reckon  £l  =  $4.44g,  or  £9  =  $40,  which  is  a  convenient  relation; 
and,  consider  the  difference  between  $4.86  and  $4.44$  (41|ct.,  = 
OTJ^J  <f0  of  $4.44|),  as  part  of  the  rate  of  exchange. 

Hence,  sterling  money  is  actually  par,  (£1  =  $4.86),  when  it  is 
quoted  9-$$%  premium,  (on  £1  =  $4.44$) ;  it  is  actually  below  par, 
when  it  is  quoted  less  than  927jj  <f0  premium,  .and  above  par,  only 
when  it  is  above  927a  %  premium. 

ART.  333.    TABLE  OP  FOREIGN  COINS  AND  MONIES. 

Amsterdam  and  Antwerp. — 1  florin  =  100  cents  =  $.40.  Some- 
times Flemish  money  is  used  as  follows  :  1  pound  =  6  florins  = 
20  schillings  =  120  stivers  =  240  groats  =  1920  pfennings. 

Bombay. — 1  rupee  =  4  quarters  =  400  reas  =  16  annas  =  50 
pice  =  45  ct.  (ct  in  this  table  means- Cents,  U.  S.  Money.) 

Brazil. — Same  as  Lisbon. 

Bremen. — 1  thaler  =  72  grootes  =  360  swares  =  78|  ct. 

Cadiz. — 1  peso  duro  =  lOf  reals  of  old  plate;  1  real  =  16 
quintas  =  34  maravedis  =  10  ct ;  1  ducat  of  plate  =  11  reals  ;  1 
real  vellon  =  5  ct 

Calcutta. — 1  gold  mohar  =  16  sicca  rupees  =  64  cahauns  — - 
256  annas  =  3072  pice  =  20480  gundas  ;  1  sicca  rupee  =  50  ct  ; 
1  current  rupee  =  44^  ct ;  a  lac  =  100000  rupeod ;  a  croro  =  100 
lacs. 

REVIEW.  —  332.  What  is  the  course  of  exchange?  Does  it  fluc- 
tuate ?  Why  ?  When  the  course  of  exchange  is  given,  how  is  the  com- 
putation performed  ?  How,  when  the  rate  of  exchange  is  given  ?  On 
which  country  is  the  rate  of  exchange  given  ?  On  which  the  course  of 
exchange  ?  What  is  the  actual  par  of  exchange  between  the  United 
States  and  England  ?  What  was  the  old  par  ?  Which  is  used  in  com- 
putations of  sterling  exchange  ?  Why  ?  What  rato  of  exchange  in  favor 
of  England  makes  sterling  money  really  par?  When  is  it  at  a  real  pre- 
mium ?  When  at  a  real  discount  ? 


EXCHANGE.  275 


Canton.— I  tael=10  mace=lOO  candarines=1000  cash=$1.48. 

Civita  VeccJiia.—l  scudo  =  10  paoli  =  100  bajocchi  =  $1. 

Constantinople. — 1  piastre  =  40  paras  =  120  aspers  =  4  ct. 

Copenhagen. — 1  rix  dollar  =  6  marcs  =  96  skillings  ;  1  rigs- 
bank  dollar  =  52  ct.;  the  old  rix  dollar  =  $1.05. 

Dantzic. — 1  thaler  =  30  silver  groschen  =  360  pfennings  = 
69  ct. ;  sometimes  the  following  are  used :  1  rix  dollar  =  3  florina 
=  90  groschen  =  270  schillings  =  1G20  pfennings  =  69  ct. 

Genoa. — 1  lira  =  100  centesimi  =  $1.86.  Old  divisions :  1  lira 
=  20  soldi ;  1  soldo  =  12  denari. 

Hamburg. — 1  mark  banco  =  16  sols  or  schillings  lubs  =  192 
pfennings  Tubs  =  $.35.  Also  1  pound  =  2j  crowns  =  3f  thalers 
=  7^  marks  =  20  schillings,  Flem.  =  240  grotes,  Flem.  Lubs  is 
a  contraction  for  money  of  Lubec.  Money  is  either  banco  or  cur- 
rent ;  the  former  being  at  a  premium  of  23  fo  over  the  latter. 

Havana. — 1  dollar  =  8  reals  plate  =  20  reals  vellon  =  $1.  A 
doubloon  =  $17. 

La,  Guayra. — 1  dollar  =  8  reals  =  75  ct. 

Leghorn. — 1  pezza  of  8  reals  =  5|  lire  =  20  soldi  =  240  denari 
=  87  ct.  The  Tuscan  corona  or  scudo  =  $1.05.  Also,  1  lire  = 
20  soldi  =  240  denari. 

Lisbon. — 1  milree  ==  1000  rees  =  $1.12  ;  1  milree  of  Azores  = 
83 3  ct ;  1  milree  of  Madeira  =  $1 ;  1  old  crusado  =  400  rees ;  1 
new  crusado  =  480  rees  ;  1  testoon  =  100  rees. 

Madras. — 1  pagoda=32  rupees=42  fanams=3360  cash=$1.84. 

Na2)les. — 1  ducato  di  regno  =  10  carlini  =  100  grani  =  $.80  ; 
1  scudo  =  12  carlini  =  $.95. 

Palermo. — 1  ducato  =  10  piccioli  =  100  bajocchi  =•  $.80.  Ac- 
counts are  generally  kept  in  the  following :  1  oncia  =  30  tari  = 
600  grani  =  3  ducati  =  $2.40. 

St.  Peiersburgh. — 1  bank  rouble  =  100  copecks  =  $.214.  Also, 
1  silver  rouble  =  360  copecks  =  $.75. 

Stockholm. — 1  rix  dollar  banco  =  48  skillings  =  576  rundstycks 
=  $.40;  1  silver  rix  dollar  =  $1.06. 

Trieste. — 1  florin  =  60  kreutzers  =  240  pfennings  =  $.48^. 

Venice.— I  lira  =  100  centesimi  =  1000  millesimi  =  $.16. 

REM. — The  values  of  the  foreign  coins  in  this  table  are  expressed 
in  the  U.  S.  gold  and  silver  coins  as  they  were  previous  to  1853. 
If  their  values  in  the  new  silver  coinage  since  1853  are  required,  the 
values  given  above  most  be  increased  7§|  per  cent.  (See  Art.  204.) 

ART.  334.    HOME,  or  INLAND  EXCHANGE. 

1.  "What  is  paid  for  $3805.40  sight  exchange  on  Bos- 
ton, at  ±%  premium?  Ans.  $3824.43 

2.  What  for  a  30  da.  bill  on  New  Orleans  for  $7216.85, 
»t  1%  discount,  interest  off  at  6  %  ?       Ans.  $7150.09 

wterest  and  discount  are  calculated  on  the  face  of  the  draft. 


276  RAY'S   HIGHER   ARITHMETIC. 


3.  What  cost  a  check  on  St.  Louis  for  $1505.40,   ai 
\%  discount?  Am.  $1501.04 

4.  What  cost  a  GO  da.  draft  on  New  York  for  $12692.50, 
at  -|  %  premium,  int.  off  at  6  %  ?  Am.  $12054.42 

5.  What  must  be  tho  face  of  a  draft  to  cost  $2000,  a* 
*%  premium?  Am.  $1987.58 

Assume  $1  for  the  face,  determine  the  cost  on  this  supposition, 
and  divide  the  given  cost  by  it. 

6.  What  must  be  the  face  of  a  draft  to  cost  $4681.25 
at  \\%  discount?  Ans.  $4740.51 

7.  Of  an  18  da.  draft,  costing  $5264.15,  at  £  %  pre 
mium,  int.  off  at  6  %  ?  Ans.  $5256.27 

Assume  $1  for  the  face,  &c.     (See  Ex.  5.)     Prove  by  selling  tht 
draft  as  proposed,  and  see  if  it  yields  the  given  amount. 

8.  Of  a  21  da.  draft,  costing  $6836.75,  at  1%  discount, 
and  int.  off  at  6  $?  Ans.  $6925.04 

ART.  335.          FOREIGN  EXCHANGE. 

1.  What  is  the  cost  in  New  York,  of  a  draft  on  London 
for  £2748  11s.  6d.,  at  10  %  prem.?     Ans.  $13437.48 

SUGGESTION. — Express  the  face  in  pounds,  reduce  to  $,  at  the 
rate  of  £1  =  $4u°,  increasing  the  result  by  10  %. 

2.  In  Liverpool,  of  a  draft  on  Boston  for  $26550,  at 
9s  %  premium  for  sterling?  Ans.  £5455  9s.  7d. 

As  £1  =  $4^  X  ~,  divide  $2G550  by  this  value,  or  multiply  it  by 
y        iuo 

9        100 
the  value  $1  =£-  X  ^— ;  reduce  the  decimal  to  shillings  and  pence. 

3.  What  is  the  face  of  a  draft  on  London,  at  9?  %  pre- 
mium, costing  $6244.50?  ^tns.  £1280  3s.  lOAd. 

Assume  £1  for  the  face,  reduce  it  to  $  at  9|  %  premium  on  $47j° 
and  divide  $6244.50  by  it 

4.  Of  a  draft  on  Philadelphia  costing  £1500,  at  91  % 
premium  for  sterling?  Ans.  $7283.33 

5.  What  is  the  cost  of  a  draft  on  Paris  for  6072. 25f., 
»fc|l«=5.85f.?  Ans.  $1135. 

6.  Of  a  draft  on  New  York,  at  Havre,  for  $2918.50, 
at  $1  =  5.37if.?  AM.  15687. 26f. 


EXCHANGE. 


7.  What  is  the  face  of  a  draft  on  Bordeaux,  to  cosf 
$453-1.20,  at  $1  =  5.40f.?  Ans.  24484.081*. 

8.  What  cost  a  draft  on  Hamburgh,  for  1231  marks  10 
schil.  8  pfen.,  at  1  mark  banco  =  33  ct.?   Ans.  $400.t45 

9.  What  cost  at  Hamburgh,  a  draft  for  $1826.  (TO.  at 
1  mark  bonco  =  35  ct.? 

Ans.  5219  marks  banco  2  schil.  3?  pfen. 

10.  A.  draft  on  Amsterdam,  for  3422  florins  15  cents, 
at  1  florin  =  38  ct,  U.  S.?  Ans.  $1300.42 

11.  Wiat  cost  at  Amsterdam,  a  draft  for  $0312.80,  at 
1  florin  =  42  ct.  U.  S.?        Ans.  15030  florins  48  cents. 

12.  What  cost  a  draft  on  Lisbon,  for  724  railrccs  960 
rees,  irt  1  milrce  =  $1.25  ?  Ans.  $906.20 

1  3.  What  cost  at  Lisbon,  a  draft  on  U.  S.  for  $3542  .  60, 
at  1  milrce  =  $1.28?  Ans.  2767  milrces  656  rccs. 

1-1  A  draft  on  Cadiz  for  1252  reals  27  maravedis  plate, 
at  1  piastre  of  8  reals  =  66  ct.?  Ans.  $103.36 

}5.  What  cost  at  Barcelona,  a  draft  for  $7564.12,  at 
I  real  =  83  ct.?  Ans.  88989  reals  22  maravedis. 

16.  What   cost   at   Stockholm,   a   draft    on    U.    S.   for 
$0529.50,  at  1  rix  dollar  =  $1.01? 

Ans.  3494  rix  dollars  26  skillings  7  rundstycks. 

17.  What  cost  a  draft  on  Stockholm,  for  5643  rix  dol. 
V  skil.  4rund.,  at  1  rix  dol.  =  98  ct.?    Ans.  $5530.29 

18.  What  cost  a  draft  on    Moscow,  for  8751  roubles 
63  copecks,  at  1  rouble  =  27  ct.?  Ans.  $2362.94 

19.  What  cost  at  Archangel,  a  draft  for  $3387.06,  at 
I  rouble  =  24  ct?         Ans.  14112  roubles  75  copecks. 

20.  What  cost  a  draft  on  Berlin,  for  634  rix  dol.  13  sil- 
ver groschcn  5  pfen.,  at  Irix  dol.  =  65  ct.?  -4ns.  $412.39 

21.  What  cost  at  Dantzic,  a  draft  for  $4687.20  at  1  rix 
dol.  =  66  ct.?     Ans.  7101  rix  dol.  24  silver  gro.  7  pfen. 

22.  What  cost  a  draft  on  Copenhagen,  for  2934  rigs 
bank  dol.  5  marks  14  skill,  at  1  rigsbank  dol.  =  48  ct.? 

Ans.  $1408.79 

23.  What  cost  at  Copenhagen,  a  draft  for  $4427.35 
at  1  rigsbank  dol.  =  51  ct.  ? 

Ans.  8681  rigsbank  dol.  T*  skil. 

24.  What  cost  at  Canton,  a  draft  for  $13653.85,  at 
I  tael  =  $1.55? 

Ans.  8808  taels  9  mace  3  candarines  5  cash. 


278  RAY'S   HIGHER   ARITHMETIC. 


ARBITKATION    OF    EXCHANGE. 

ART.  336.    Exchange  may  be  either  direct  or  circular. 

Direct  exchange  is  confined  to  two  places,  the  residence 
cf  the  drawer  and  of  the  drawee. 

Circular  exchange  is  effected  by  a  succession  of  bills 
drawn  a^  and  on,  one  or  more  intermediate  points. 

The  circular  method  is  called  Arbitration  of-  Exchange. 

A  New  York  merchant  wishes  to  remit  $2240  to  Lisbon: 
is  it  more  profitable  to  buy  a  bill  directly  on  Lisbon  at 
1  milree  =  $1.15:  or,  to  remit  through  London  at  8  % 
premium,  and  through  Paris  at  £1  =  25.20f.,  to  Lisbon 
at  1  milree  =  6f.? 

SOLUTION. — The  direct  exchange  gives  the  value  of  §2240  = 
1947.826  milrees,  by  dividing  $2240  by  $1.15,  the  value  of  1  milree 
in  exchange.  The  value  of  the  remittance  by  the  circular  exchange 
is,  2240  X  ^Ox  l  Q8  X  25.20  X  g  =  I960  railrees.  The  $2240  is 
reduced  to  £,  by  multiplying  it  by  the  value  of  $1.  which  is 
£-|  ft  '  the  £  are  reduced  to  francs,  by  multiplying  by  the 

value  of  £1,  which  is  25.20f.;  the  francs  are  reduced  to  milrees,  by 
multiplying  by  the  value  of  1  franc,  which  is  g  milree.  After  in- 
dicating the  operations,  cancel. 

2d  SOL.— The  same        £0     $£2^0      =      (     )  milreeB. 
result  can  be  obtained         28      1  milree      =      fif.  2 
by  a  chain  of  equations,  7     2  5 . 2  0  f.     =      £1 

each     expressing     the  £<•)  =      $-4  Q 

course  of  exchange  be-  ft  X 

tween  two  points  in  the       .09)176.40  J-P? 

circuit,  except  the  first,      G.       — iggo^ 
which  has  the  sum  to      p.^  ex>  }  9  4  7 .  g  £  f.       <  0  9 

be  remitted  equal  to  an  ' ! L_  .  - 

unknown  number  (    )  Gain  1  2,1  7'*  Jnilrsea. 

of  the  denomination  re- 
quired (milrees);  these  equations  are  so  arranged  that  the  same  de- 
nominations appear  on  both  sides  of  the  sign  (  =  );  the  product  of 
both  sides  being  equal,  the  vacant  term  id  found  by  dividing  the 
product  of  those  on  the  other  side,  by  the  product  of  the  others  on  ita 
-wn  side,  canceling  first,  to  shorten  the  v.ork. 


REVIEW.  —  336.    What  two  kinds  of  ej-hange?     What  is  direct 
Bhange?     Circular?     What  is  the  circular  c.-Jled  ?     Solve  the  ezarap.o- 


ARBITRATION   OF   EXCHANGE.  279 


ART.  337.  Hence,  to  compare  two  quantities  of  different 
denominations,  through  the  medium  of  other  denomina- 
tions, this  rule,  called  the 

CHAIN  RULE. 

RULE.  I. — Reduce  the  given  quantity  to  the  denomination  with 
which  it  is  compared;  reduce  this  result  to  the  denomination  with 
which  it  is  compared;  and  so  on  until  the  required  denomination 
is  reached,  indicating  the  operations  by  writing  as  multipliers,  the 
proper  unit  values.  The  compound  fraction,  thus  obtained,  re- 
duced to  its  simplest  form,  will  be  the  amount  of  the  required 
denomination. 

ROLE  II. — Form  a  series  of  equations,  expressing  the  conditions 
of  the  question ;  tlie  first  containing  the  quantity  given  equal  to 
an  unknown  number  of  the  quantity  required,  and  all  arranged 
in  such  a  way  that  the  right  hand  quantity  of  each  equation  and 
the  left  hand  quantity  in  the  equation  next  following,  shall  be  of 
the  same  denomination,  and  also  the  right  hand  quantity  of  the 
last  and  the  left  hand  quantity  of  the  first.  Cancel  all  equal 
factors  on  the  right  and  left,  and  divide  the  product  of  the  quan- 
tities in  the  column  which  is  complete  by  the  product  of  those  in 
the  other  column.  The  quotient  will  be  the  quantity  required. 

NOTES. — 1.  This  is  called  the  chain  rule,  because  each  equation 
and  the  one  following,  as  well  as  the  last  and  first,  are  connected 
like  a  chain,  having  the  right  hand  quantity  in  one  of  the  same 
denomination  as  the  left  hand  quantity  of  the  next. 

2.  In    applying  this  rule   to  exchange,   if  any  currency  is  at  a 
premium  or  a  discount,  its  unit  value  in  the  currency  with  which 
it  is  compared,  must  be  multiplied  by  such  a  number  of  decimal  hun- 
dredths,  as  will  increase  or  diminish  it  accordingly.    (See  Operation.) 

3.  If  commission  or  brokerage  is  charged  at  any  point  of  the  cir- 
cuit, for  effecting  the  exchange  on  the  next  point,  the  sum  to  be 
transmitted  must  be  diminished  accordingly,  by  multiplying  it  by 
the  proper  number  of  decimal  hundredths. 

EXAMPLES  FOR  PRACTICE. 

1.    A  New  York  merchant  remits  $1460  to  Hamburgh, 
through  London  and  Paris:  what  is  the  value  when  received, 
if  £1  =  $4.85;  25.20f.  =  £1;  1  mark  banco  =  l.SOf.? 
Ans.  4214  mark  bancos  6  sols  11  pfennings. 


REVIEW. — 337.  What  is  the  chain  rule?  Why  is  this  called  the  chain 
rule?  AVliy  should  this  chain  or  connection  be  made?  If  any  of  the  cur- 
rencies is  at  a  premium  or  discount,  what  should  be  done?  If  commission 
or  brokerage  is  charged  at  any  point  of  the  circuit,  what  should  be  done  ? 


•280  RAY'S  HIGHER  ARITHMETIC. 

2.  A,  of  Philadelphia,  wishes    to   remit  $4532.80   to 
St.  Petcrsburgh :  the  direct  exchange  is  $1  =  3  roubles  3G 
copecks,  while  through  London,  Paris,  and  Dantzic,  it  ia 
as  follows:  8  %  in  favor  of  London;  25.20f.  =  £1 ;  1  th.-iler 
=  4f. ;  '2  roubles  GO  copecks  =  1  thaler.     Which  is  better? 

Ans.  Circular,  by  509  roubles  94  copecks. 

3.  A,  of  Boston, owes  in  Amsterdam  10000  florins.    How 
much  in  U.  S.  money  will  pay  it,  by  remitting  through  Lon- 
don and  Paris  at  the  following  rates:  9%  in  favor  of  London; 
£1  =  25f. ;   1  florin  =  2.32f. ;  allowing  1  %  brokerage  in 
London,  and  3  %  in  Paris?  Ans.  $4563.87 

SUGGESTION. — Multiply  the  £  by  .99,  and  the  francs  by  .OO^,  to 
allow  for  the  brokerage.  (Note  3.) 

4.  A,  of  Memphis,  has  $6000  to  pay  in  New  York.    The 
direct  exchange  is  1  %  premium;  but  exchange  on  New 
Orleans  is  k.-%  premium,  and  from  New  Orleans  to  New 
York  is  1  %  discount.     By  circular  exchange,  how  much 
will  pay  his  debt,  and  what  is  his  gain  ? 

Ans.  $5969.70;  $90.30  gain. 

SUGGESTION. — In  home  exchange,  the  currency  being  the  same 
throughout  the  circuit,  annex  to  the  $,  the  name  of  the  place,  as 
S  N.  Y,  or  $  N.  0.,  for  New  York  money  and  New  Orleans  money. 

5.  A  merchant  of  St.  Louis  wishes  to  remit  $7165.80 
to  Baltimore.    Exchange  on  Baltimore  is  \%  premium;  but, 
on  New  Orleans  it  is  3  %  premium;  from  New  Orleans  to 
Havana,  i  %  discount;   from  Havana   to   Baltimore,  1  </0 
discount.     What  will   be   the  value   in  Baltimore  by  each 
method,  and  how  much  better  is  the  circular? 

Ans.  Direct,  $7112.46;  circ.,  $7238.36;  gain,  $125.90 

6.  A    Louisville   merchant    has    $10000    due   him   in 
Charleston.     Exchange  on    Charleston    is  :{-  %   premium. 
Instead  of  drawing  directly,  he  advises  his  debtor  to  remit 
to  his  agent  in  New  York  at  g  %  premium,  on  whom  he 
immediately  draws  at  12  da.,  and  sells  the  bill  at  IA  c/0 
premium,  interest  off  at  6  %.    What  does  he  realize  in  this 
way,  and  what  gain  over  the  direct  exchange? 

Ans.  Realizes  $10087.17;  gain,  $62.17 
SUGGESTION. — Interest  nt  6  %  per  annum  is  A  %  a  month,  or  \  % 
for  \~>  days,  "which  diminishes  the  rate  of  premium  to  \\  fa- 

7.  A  Cincinnati  manufacturer  receives,  April  18th,  an 
account  of  sales  from  N.  Orleans  ;  net  proceeds  $5284.67, 


ARBITRATION   OF   EXCHANGE.  28\ 

due  June  4|7.  He  advises  his  agent  to  discount  the  debt  at 
6  %,  and  invest  the  proceeds  in  a  7  da.  bill  on  New  York, 
interest  off  at  6  %,  at  1A  %  discount,  and  remit  it  to  Cin- 
cinnati. The  agent  does  this,  April  26.  The  bill  reaches 
Cincinnati  May  3,  and  is  sold  at  5  $  prcm.  What  is  the 
proceeds,  and  how  much  greater  than  if  a  bill  had  been 
drawn  May  3,  on  New  Orleans,  due  June  7,  and  sold  at 
|  %  prern.,  and  int.  off  at  6  %? 

AM.  Proceeds,  $5383.32;  gain,  $109.66 

8.  Find  the  net  value  in  Cincinnati,  of  a  sum  due  in 
Paris,  of  12GOOf.,  which  is  transmitted  to  London,  at  1  % 
commission;  exchange,  25.20f.  =  £1;  a  draft  for  the  net 
amount  in  London,  is  drawn  at  New  York  at  7  %  premium, 
after  deducting  ^  $,  commission;  the  proceeds  of  this  bill 
being  remitted  to  Cincinnati,  in  a  check  at  A  %  discount. 

"An».  $2354. 

9.  If  5  oranges  are  worth  8  lemons,  3  lemons  worth 
10  apples,  4  apples  worth  1  melon,  and   6  melons  worth 
75  ct.;  how  much  money,  are  12  oranges  worth?  Ans.  $2. 

10.  If  3  men  do  as  much  as  7  women,  and  10  women 
as  much  as  27  boys,  and  42  boys  as  much  as  75  girls,  and 
36  girls  can  bind  500  sheaves  in  an  hour,  how  many  can 
12  men  bind  in  an  hour?  -4ns.  1875. 

11.  What  must  9  oz.  of  tea  be  sold  for,  if  45  Ib.  cost 
$28,  and  the  retail  profit  is  20  %1  Ans.  42  ct. 

12.  What  should  coffee  be  sold  at  per  pound,  if  an  in- 
voice of  253  bags,  averaging  166;?  Ib.  each,  cost  $4620,  the 
freight  and  other  charges  being  15  %  on  the  invoice,  and 
the  profit  on  the  whole  cost,  25  %?  Ans.  15|  ct. 

13.  If  the  freight  at  New  Orleans,  per  U.  S.  bu.  of  corn 
(56  lb.)>  is  10d.,  what  is  the  freight  at  Liverpool,  per  impe- 
rial quarter  (480  Ib.) ;  primage,  5  %  ?  Ans.  7s.  6d. 

REMARK. — Primage  is  an  allowance  to  the  master  and  mariners 
of  a  ship  for  loading,  and  is  charged  on  the  freight. 

14.  What  is  the  freight  on  a  cargo  of  2560  bales  of 
cotton,  averaging  450  Ib.  each,  at  j|d.  per  pound,  allowing 
5  %  primage,  8  %  premium  on  sterling  money,  and  5  <f0 
commission  for  advancing  the  freight?       Ans.  $23814. 

15.  If  the  freight  per  U.  S.  bu.  of  corn  (56  Ib.),  is  26  ct., 
charges  4  ct.  per  bu.  (as  usual  in  New  Orleans),  making 
the  freight  30  ct.  per  bu.  on  board,  what  is  the  freight  per 

24 


RAY'S   HIGHER   ARITHMETIC. 


imperial  quarter  (480  lb.),  allowing  4  %  commission   fo? 
advancing  the  freight,  sterling  exchange  4  %  premium  ? 

Ans.  11s.  Cfd. 

16.  If  the  freight  is  66  ct.  per  U.  S.  bu.,  charges  and 
commission  as  in  last  example,  exchange  8  %  prem.,  what 
is  the  freight  per  imperial  quarter?  Ans.  26s. 

17.  What  is  the  cost  in  Liverpool,  of  corn  per  quarter, 
ought  in  New  Orleans  at  5 Oct.  per  U.  S.  bu.,  charges  4ct. 

per  bu.,  exchange  8  %  premium,  freight  13d.  per  bu.,  com- 
mission (on  Freight  only^)  5  %  ?  Ans.  29s.  |d. 

Find  the  cost  per  qr.;  then,  the  freight  per  qr.,  and  add. 

18.  At  how  many  cents  per  U.  S.  bushel  can  an  order 
be   filled   in    New   Orleans,   limited   at   28s.  per  imperial 
quarter,  the  freight  being  15d.  per  U.  S.  bu.,  commission 
5%*,  and  exchange  6  %  premium?  Ans.  42  ct. 

Find  the  freight  per  imperial  qr.,  subtract  it  from  28s.,  and  find 
the  cost  per  U.  S.  bu.  corresponding  to  the  remainder.  This  cost, 
less  4ct.  the  usual  charges  per  bu.,  is  the  price  per  U.  S.  bu. 

19.  How  many  cents  can  be  paid  for  corn  in  New  Or- 
leansr  to  fill  an  order  from  Liverpool,  limited  to  80s.  per 
imperial  quarter,  the  freight  being  lOd.  per  U.  S.  bu.,  com- 
mission* 5  %,  exchange  7  %  premium?       Ans.  6875  ct. 

*In  Examples  17, 18  and  19,  Commission  is  charged  on  Freight  only. 


XXVII.   ACCOUNTS  CURRENT. 

ART.  338.  THE  account  which  a  banker  keeps  of  money 
received  and  paid  out,  is  an  Account  Current. 

Accounts  Current  are  also  kept  by  merchants,  showing 
the  value  of  goods  delivered  to  their  agents  and  customers, 
and  the  payments  made  thereon. 

Hero  is  an  Account  Current  of  a  bank  with  a  deposi- 
tor : 

REVIEW. — 338.  What  is  an  account  current?  By  whom  kept?  What 
does  the  Dr.  side  show?  Tho  Cr.  side?  What  is  settling  an  account  cur- 
rent ?  What  is  it  also  called?  What  ii  necessary  in  making  the  balance  1 
What  is  the  direct  rule? 


ACCOUNTS  CURRENT. 


283 


Da. 


Henry  Armor,  in  acc't  with   City  Bank. 


CK. 


1S55. 

$ 

ct. 

1864. 

$  let 

Prod. 

Jan 

6 

To  check, 

300 

Dec. 

31 

By  bal.  old  acc't, 

(00 

2500 

H 

2<t 

>      •• 

500 

1855 

200 

400 

ii 

21 

!»          » 

100 

Jan. 

7 

»  cwh, 

60 

>i 

27 

»          If 

860 

,, 

15 

260 

2000 

9| 

3i 

>l          » 

75 

»       » 

400 

650 

8250 

» 

24 

150 

150 

» 

19 

60 

150 

•i        ii 

1000 

1050 

8150 

200 

800 

Bal.  items,    125 
Bal.  int^           2.07 

12.400 

Total  bal.  "$127.07 

$2.07" 

The  left  hand,  or  Dr.  side,  shows,  with  their  dates,  the 
sums  paid  by  the  bank  on  the  checks  of  Henry  Armor,  for 
which  he  is  their  debtor. 

The  right  hand,  or  Cr.  side,  shows,  with  their  dates,  the 
sums  deposited  in  the  bank  by  Henry  Armor,  for  which  he 
is  their  creditor. 

Settling  or  closing  an  account,  is  finding  how  much  is  due 
at  a  particular  time,  between  the  parties.  It  is  sometimes 
called  striking  a  balance.  * 

Generally,  in  an  account  current,  each  item  draws  in- 
terest from  its  date  to  the  day  of  settlement ;  hence, 

TO   SETTLE   AN   ACCOUNT   DRAWING   INTEREST. 

RULE. — Find  the  sum  of  (lie  items  on  the  Dr.  side,  with  iht 
interest  of  each  from  its  date  to  the  date  of  settlement ;  do  lh* 
same  for  the  Cr.  side.  The  difference  of  these  sums  will  be  th* 
balance  in  favor  of  the  greater  side. 

The  following  is  a  more  convenient  and 

PRACTICAL  RULE. 

Multiply  the  1st  item  by  the  number  of  days  from  the  1st  to 
the  2d  dates.  Find  the  balance  after  the  2d  item,  and  multiply 
it  by  the  number  of  days  from  tfie  2d  to  the  '3d  dates,  and  to 
on  to  the  last  balance,  which  is  multiplied  by  the  number  oj 
days  from  the  last  date  to  the  date  of  settlement. 

Each  product  is  put  into  a  Dr.  or  Cr.  column,  according  a» 
the  balance,  from  which  it  comes,  is  Dr.  or  Cr.  Balance  the 


R  E  v  i  E  Vf. — 338.  What  is  the  practical  rule  ?    Show  the  reason  o 
rule  as  applied  to  the  account  given. 


284  RAY'S   HIGHER   ARITHMETIC. 

two  columns  of  products,  and  find  the  interest  on  this  balanct 
Jbr  1  day,  at  the  at/ reed  rale.  This  will  be  the  balance  of  in 
tfrent  in  Jit  cur  of  the  larger  column,  which  being  added  to,  or 
subtracted  J'rom,  the  last  balance  of  Hems,  according  as  they 
are  on  the  same  or  opposite  sides,  will  give  the  final  balance  re- 
quired. 

DKMO.NSTRATIOX. —  Take   the   account  already  given.     The    Is 
item,  S50I),  being  a  balance  in  favor  of  Henry  Armor  from  an  ol 
account,  is  on  interest  from  its  date  (Dec.  31,  1854)  to  the  date  of 
the  next  item  (Jan.  5),  which  is  5  clays.     The  interest  of  $500  for 
6  da.  is  the  same  as  the  interest  of  $-300  X  &  =  $-500  for  1  da. 

Place  $2500  in  the  column  of  products  opposite;  find  the  balance 
of  items  ($200),  which  is  the  difference  between  the  sum  on  deposit 
(§500),  and  the  sum  ($300)  paid  out  on  the  check.  This  balance 
may,  for  convenience'  sake,  be  set  down  in  pencil  or  red  ink  on  the 
eide  of  the  account  to  which  it  belongs,  as  shown  in  the  form.  This 
balance  is  on  interest  from  the  second  date  (Jan.  5)  to  the  third 
date  (Jan.  7),  =  2  days  ;  the  interest  of  $200  for  2  da.  is  the  same 
as  the  interest  of  $200  X  2  =  $400  for  1  da, ;  set  this  product 
$400  in  the  column  opposite,  and  keep  on,  taking  a  balance  and 
forming  a  corresponding  product^  for  every  item,  to  the  day  of 
settlement. 

In  this  example,  the  balance  being  always  on  the  Cr.  side,  the 
products  will  all  be  in  one  column;  but,  if  the  balance  were  some- 
times on  one  side,  and  sometimes  on  the  other,  a  Dr.,  as  well  as  a 
Cr.  column  for  products,  would  be  needed  ;  adding  each  column,  the 
difference  of  the  sums  would  be  the  balance  in  favor  of  the  larger 
column ;  the  interest  of  this  difference  for  Ida.,  would  be  the  excess 
of-  interest  in  favor  of  the  larger  column. 

The  balance  of  products  in  this  example,  is  $12400;  its  interest 
for  CO  da.  at  (>%  is  $124.00  (Art'.  303) ;  dividing  by  GO  gives  the  in- 
terest for  1  da.,  $2.07,  in  favor  of  the  Cr.  side  :  as  the  balance  of 
items,  $125,  is  also  in  favor  of  the  Cr.  side,  the  total  balance  to  Henry 
Armor's  credit,  is  the  sum  of  the  two,  ($127.07). 

NOTE.  —  In  settling  depositors'  accounts  in  New  York,  Great 
Britain,  &o.,  the  interest,  as  ordinarily  obtained,  must  be  dimin- 
ished by  ^'.j  of  itself,  (Art.  308) ;  but  elsewhere,  this  deduction  should 
not  be  made  ;  as,  there  is  no  more  reason  for  it  in  case  of  depositors' 
accounts,  than  in  the  discounts  of  the  bank:  it  is  manifestly  unjust 
for  the  bank  to  make  a  deduction  on  the  interest  it  pays,  and  not 


REVIEW.  —  338.  When  will  two  columns  for  products  be  necessary? 
Uow  do  we  then  proceed?  On  which  side  will  the  balance  of  interest  be? 
\Vhat  should  be  done  with  it  when  obtained  T 


ACCOUNTS   CURRENT. 


285 


to  allow  it  on  the  interest  it  receives,  the  circumstances  being 
precisely  the  same :  if  it  insists  on  taking  the  time  by  days  in  its 
discounts,  it  ought  not  to  object  to  any  advantage  to  the  depositor 
in  making  up  his  interest  on  the  same  plan.  It  is  true,  it'  the  in- 
terest on  a  depositor's  account  were  calculated  on  each  item  sepa- 
rately to  the  time  of  settlement,  using  calendar  months  when 
possible,  the  result  would  be  slightly  less  than  by  the  last  rule, 
a  day  being  occasionally  lost,  but  it  would  not  be  ^'3  leas :  the  dif- 
t.-reuce  would  vary  with  the  size  and  dates  of  the  items. 

FIND  TUB  INT.  DUE,  AND  BALANCE  THESE  ACCOUNTS. 

1.     Da.          F.  II.  Willis  in  acc't  with  E.  S.  Kennedy.  CR. 


1849. 

Dol. 

ct. 

1S43. 

Dol.'ct. 

Product*. 

Jaw. 

• 

To  check, 

170 

Dec. 

a 

By  bal.  forw. 

325 

u 

ii      » 

480 

1849. 

M 

1C. 

>»      » 

96 

Jan. 

7 

„   cash  dep. 

800 

» 

18 

»      >» 

600 

,, 

20 

»           11                 !> 

176 

ft 

•/8 

11      >• 

50 

•' 

30 

»      11        >i 

240 

Interest  to  Jan.  31,  at  G  per  cent. 

An*.  lut  due  Willis,  $2. 33:  bal.  due  Willis,  $246.33 
*L     DR.      A.  L.  Morris  in  acc'l  with  T.  J.  Fisher  $  Co.  CR. 


is:,i. 

—  -            - 

Dol. 

,t. 

1853. 

Dot.  'ct. 

Product*. 

Jan. 

13 

To  cbeck. 

350 

Dec. 

::i 

By  bal.  from  old  acc't. 

813 

M 

,, 

L'_' 

275 

1854. 

Feb. 

2. 

100 

Feb. 

4 

„  cash, 

12" 

Mny. 

1 

»      » 

400 

Mar. 

17 

„     „ 

500 

Jun. 

83 

«>      n 

108 

a 

May 

HI 

>»     » 

H 

BO 

Interest  to  June  30,  at  6  per  cent. 
Ans.  Int.  duo  Morris,  $13.34:  bal.  due  Morris,  $298.23 

3.     DR.  Wn.  Whi.'e  in  acc't  with  Beach  $  Berry.  CR. 


1855. 

Dol. 

Dol. 

et 

Products. 

Jul. 

I 

To  check, 

212 

50 

.'in 

30 

By  bal.  from  old  acct. 

1102 

K 

„ 

jo 

ii      ii 

Cfi 

Jul. 

6 

„  uish  dep., 

60 

Aug. 

7 

2:55 

n 

ir, 

99        if        II 

95 

H 

2J 

ii      11 

30(1 

Au_ 

y 

108 

T8 

Sep. 

;, 

ii      ii 

110 

Sep. 

IS 

ft          M          II 

32 

,, 

11 

ti      ii 

46 

M 

Oct. 

8 

79 

DO 

21 

ii      ii 

454 

28 

Interest  to  Oct.  12,  at  Id  pel  cent. 
Ans.  Int.  due  White,  §19. 68;  bal.  due  White,  $123.68 


UKVIKW.  —  338.  In  New  York,  ic.,  whal  bhould  bo  done  in  finding 
Ibe  interest?     Should  this  be  done  elsewhere?     Why  not? 


280 


RAY'S   HIGHER   ARITHMETIC. 


ART.  339.  The  rule  (Art.  338)  is  not  so  applicable 
to  a  merchant's  account  current,  where  the  items,  being 
sometimes  notes,  drafts,  acceptances,  sales  on  credit,  &e., 
are  not  generally  due  in  the  order  they  are  written. 

TO   BALANCE   A   MERCHANT'S   ACCOUNT   CURRENT, 

HULE  — Multiply  each  item  by  the  number  of  days  from  the 
ime  it  is  due  as  cash  until  the  day  of  settlement.  Set  the  pro- 
ducts of  the  Dr.  items  in  a  Dr.  column,  and  those  of  the  Or.  items 
in  a  Cr.  column. 

Add  each  of  the  columns,  and  find  the  difference  of  their 
turns;  the  interest  of  this  for  1  day,  at  the  rate  agreed,  u-ill  be 
the  balance  of  interest  in  favor  of  the  larger  column,  which  is 
placed  on  the  side  of  the  account  where  it  belongs ;  the  difference 
between  the  two  sides  will  be  the  whole  balance  in  favor  of  the 
larger  side. 

REMARK. — This  rule  serves  for  closing  an  account  sales,  when 
the  amount  due  at  a  particular  date  is  required. 


4.     DR. 


F.  Archer,  in  acc't  with  T.  $  W.  Day. 


CR. 


1855. 

Dol. 

ct. 

Products.     ISM. 

Dol. 

ct. 

Frodueti. 

Jun 

'-".I 

To  100  barrels 

!Jul. 

2(1 

By    20    hhd. 

pork,  @  $12, 

1200 

sugar,  @  $t;6 

1320 

Jnl. 

15 

To  50  barrels 

Aug 

20 

By    15    bales 

flour,  @  $9, 

450 

cotton,  COOO 

Sep. 

9 

To  40  barrel? 

lb.,  @8c., 

4SO 

whisky,  1380 

NOT 

12 

By  8  bhd.  to- 

gnl.,@ 25  c. 

340 

bacco,  12000 

Dec. 

18 

To  18  barrels 

lb.,  @  i~  -., 

720 

i 

pork,  @  $12, 

216 

1  1 

Interest  at  6  per  cent,  to  Dec.  31. 

Am.  Int.  due  T.  &  W.  Day,  $3.95:    bal.  due  Archer, 
$310.05       The  1st  rule  may  be  used  here  to  advantage. 

5.     DR.          Oliver  Mason,  in  acc't  with  Sharpe  $  Swift.         CR. 


18M.    I 

da 

Dol. 

,-t. 

Prod. 

I8U, 

da 

Dol. 

ct. 

Prod. 

Jan.    6!To  invoice,  cash 

874 

78 

Mar.    6 

By  cash, 

372 

Feb.    3 

»        »»       >» 

293 

Apr.    8 

,,      ,, 

495 

May  16 

,  acceptance, 

May    1 

„    remittance, 

May  31, 

691 

'2"i 

June  30, 

2000 

July    6 

To  invoice,  3m. 

1000 

Slay    2 

,,     remittance, 

.  Aug.  20 

„      2  m. 

1150 

II) 

May  9, 

1426 

N 

Sept.   1 

,  check, 

925 

Ant;.  15 

„    cash, 

250 

..    25 

,  cash, 

540 

Sept.  28 

,,    remittance, 

Oct.  15. 

342 

Ans.  Int.  due  Mason,  $30.73  ;  bal.  due  Sharpe  &  Swift 
$1057.87  Int.  at  6  %  to  Sept.  30. 


ACCOUNTS   CURRENT. 


287 


NOTE. — In  this  account,  two  items  on  the  Dr.  side  and  one  on  (he 
Cr.,  fall  due  after  Sept.  30,  and  must  therefore  suffer  discount,  or, 
what  is  the  same,  draw  interest  for  the  other  side :  hence,  the  products 
obtained  from  them,  are  put  on  the  side  of  the  account,  opposite  to 
that  which  contains  the  items. 

6.  Acccu*t  sales  of  600  hhd.  sugar,  per  ship  Wave,  from  New  Orleait, 
sold  by  order  and  for  account  of  1'errot,  Clarke  £  Co. 
Discount  and  interest  6  %  to  date. 


1855. 
May 

;  J'ne 

May 

J'TM 

ID 
IS 

n; 

•J5 

1 

-21} 

600  hhd.  sugar. 
To    Jonas    A    King,   2   mon.,    100   hhd., 
^107210  lb.),  @,6c  

Dol. 
6360 
7946 
4695 

10460 
28462 

4734 

ct. 

50 
M 

57 

12 

69 

88 

Dug 
Jul.10 
AuglS 
Sep.16 

Sep.  25 

May  1 

Da 

14 

68 

88 

91 

r,6 
r,0) 

Prod'cti. 
75047 
4211G5 
385037 
951871 

184072 

To  11.  Johnson,  3  mon.,  150  hhd.  (158930 
lb  )  @  6c      •  .                

To  K.  Force,  3  mou.,  80  hhd.  (85374  lb.), 

To  K.  Woodruff,  3  mon.,  170  hhd.  (1UC184 
Jb.),  @  5ic  

Total,  800  hhd  

CHARGES. 

Freight,  600  hhd.  @  $6  $3000 

Cartage,  labor,  cooperage  287 

$3287 
Commission  2J  p.  ct.,  on  $28462.69     711.57 
Guarantee,  1      „        „        „              284.63 

20171.92 

336.199 
Interest 

1 

Discount  on  credit  sales,  and  inter- 
est on  charges,  reduced  to  date,     33G.20 

23728 

dl 

(Errors  and  omissions  excepted.) 
Ntw  YORK,  June  26,  1855. 
KDWAKO  PARKER. 

REMARK. — The  charges  paid  May  1,  are  on  interest  till  June  2G 
(56  da.);  the  sales  being  on  credit,  the  payments  are  not  due  until 
yarious  times  after  June  26,  and  must  be  discounted  for  14,  63,  82, 
and  91  days,  respectively,  to  reduce  them  to  cash  on  June  26. 


7.     Da.         E.  S.  Lee,  in  acc't  current  with  Burns 
•Interest  9  f0  to  June  iJO. 


Co.        CH. 


1851.   1 

•l;l 

Dol.  ct. 

Prod. 

1851. 

da  n..!. 

ct. 

Prod. 

Feb.  10  To  invoice,  cash 

800 

Feb.  20 

By  remit.  Mar.  6 

600 

Mar.  25 

,i       >i          i> 

900 

Mar.18 

„      „       Apr.  10 

400 

Apr.  15 

..       .,    2  mon. 

700 

Apr.  5 

.,  cash, 

200 

May  18 

..       ..    2  mon. 

600 

May  20 

,,     „ 

300 

May  31 

„    acceptance, 

Jun.20 

„  remit.  July  15 

1600 

15da.  sight, 

400 

June  20 

,,  invoice,  cash 

1000 

Ans.  Int.  duo  Burns  &  Co.,  $27.18  ;  bal.  due  Burns  & 
Co.,  $1227.18       (*  Days  of  grace  are  not  counted.) 

REVIEW. — 339.   Why  is  the  rule  for  bank  accounts  not  so  applicable  t* 
ft  merchant's  account  current?     What  is  the  rule  for  it? 


•088 


RAY'S   HIGHER   ARITHMETIC. 


8.     Da.         Joshua  Parkins,  in  acc't  with  II.  K.  Foster. 
*  Closed  Dec.  31 ;  interest  6  %. 


CR. 


F  i  

fl 

ISM.    1 

da 

Dot. 

ot. 

Prod. 

1853.   | 

da 

Dol. 

ct.  Prod. 

Jan.  15'To  invoice  3m. 

738 

Feb.    6  By  remittance, 

Ke.b.  2<i 

„    ii"c-cptance, 

Feb.  18 

250 

!.">  da.  Huhl. 

4CO 

Mar.  15 

„   remittance, 

Apr.  15 

invoice,  3  m. 

650 

15 

Bight, 

380 

June   1 

„      2m. 

G32 

May  10 

„  aw't  sales, 

•_>!.->  :'.o 

.,  Aug.  31 

acceptance, 

Aug.  10 

,,   cash, 

•A-16  15 

:!0  da.  fight. 

920 

&'t 

Oct.     S» 

„      ,. 

m 

Sept.  10 

invoice,  cash 

240 

Oct.  20 

„  remittance, 

Oct.    1!> 

.,      2  in. 

184 

•21 

Nov.  14, 

203 

HO 

I>ec.  21 

draft,  light, 

100 

Dec.    1 

„  remittance, 

| 

Dec.  15. 

500 

Ans.  Int.  due  Foster,  $49.04;  bal.  due  Foster,  $1754.91 

*  In  these  AcH.  current,  and  in  Equation  of  Payments,  days  of  graco 
arc  not  counted. 


STORAGE    ACCOUNTS. 

ART.  340.  Storage  accounts  are  similar  to  bank  accounts, 
one  side  showing  how  many  bbl.,  packages,  £c.  are  received, 
and  at  what  times;  the  other  side  showing  how  many  have 
been  delivered,  and  at  what  times. 

Storage  is  generally  charged  at  so  much  per  month  of 
30  da.  on  each  bbl.,  package,  &c. 


TO  FIND  TUB  BALANCE  DUE   ON  A  STORAGE  ACO  T 


RULE. — Multiply  the  number  o/"  bbl.  first  set  down  by  the  number 
+f  days  from  the  1st  to  the  2d  date  of  the  account,  and  set  the 
result  in  the  column  of  products  opposite. 

Find  the  number  of  bbl.  on  hand  after  the  second  item  has  been 
get  down,  and  multiply  it  by  the  number  of  days  from  the  2d  to 
the  '3d  date,  setting  the  result  in  the  column  of  products  opposite. 

Continue  thus,  multiplying  the  number  of  bbl.  finally  on  hand, 
by  the  number  of  days  from  the  last  date  to  the  date  of  settlement 

Add  these  products,  and  divide  their  sum  by  30:  the  quotient 
will  be  the  number  of  bbl.  stored  per  month,  and  this  number 
multiplied  by  the  price  of  storage  on  each  bbl.  per  month,  will 
give  the  balance  due  on  the  account. 


RE  v i E  w.— S40.  How  is  storage  charged?    What  is  the  rulo  for  settling 
ft  storage  account  T 


EQUATION   OF   PAYMENTS. 


28» 


1.    Storage  to  Jan.  31,  at  5  ct.  a  bbl.  per  mon. 

KKCE1VKI).  DELIVERED. 


1*56. 

bbl. 

JJalance  on  h»a4. 

I*,.. 

Prodaoti. 

1856. 

bbl. 

,  J»uuitry. 

2 
6 

tfiO 
150 

January, 

10 
13 

110 
00 

7 

:io 

17 

20 

10 

120 

20 

115 

11 

80 

•25 

110 

17 

160 

27 

72 

20 

76 

:« 

100 

21 

HO 

31 

23 

1.1/0 

Ans.  Storage,  $19.25  ;  bbl.  on  hand,  418. 
2.    Storage  to  Feb.  20,  at  5  ct.  a  bbl.  per  mon. 

RECEIVED.  DELIVERED. 


r  — 

bbi. 

Balance  on  band. 

Da.yi. 

Product*. 

bbl. 

.rammrv. 

si 

418 

February. 

6 

ino 

February. 

t 

2.V) 

10 

80 

i 

12*1 

12 

220 

IS 

KM) 

14 

140 

ID 

30 

IS 

90 

20 

288 

Ans.  Storage,  $15.85;  bbl.  on  hand,  0. 


XXVIII.    EQUATION  OF  PAYMENTS. 

ART.  341.  Equation  of  Paymenf$t  or  averaging  accoiinfx, 
is  a  method  used  by  merchants,  by  which  debts  due  at  dif- 
ferent times  can  be  paid  in  one  sum,  and  at  one  time, 
without  loss  to  either  party. 

A  buys  an  invoice  of  $250  on  3  mon.  credit;  $800  on 
2  mon. ;  and  $1000  on  4  uion.;  and  gives  his  note  for  the 
whole  amount :  how  long  should  the  note  run  ? 

SOLUTION. — The  whole  debt  is  $'2050;  and  since 
the  discount  on  $250  for  3  mon.  =  discount  on    $750  for  1  mon. 
and        „         „   $800  „    2    „     =        „         „  $1000  „ 
and        „        ,,$1000  „    4    „    ==        „        „  $4000  „        „ 


The  discount  on  $-050  .     .     .     .  =  discount  on  $0350  for  1  mon. 
But  die  discount  on  $(1350  for  1  mon.  is  the  same  as  the  discount  on 
$'2050  for  §5*  A  8  mon-i  =  B-iV  niou.  =  3  tuou.  «i  da.;  which  is,  there- 


Re  ri  B  w. — 341.  What  is  equation  of  paymonta?     Solve  the  example. 
25 


RAY'S   HIGHER   ARITHMETIC. 


fore,  the  tiroi  the  note  should  run,  since  it  IB  the  time  for  which  the 
debt,  ($'2060),  must  be  Jiscounted  to  be  considered  cash.  The  time 
thus  fouud  is  called  the  mean  or  average  time. 

I  buy  of  a  •whulcsalc  dealer,  at,  3  mon.  credit,  as  follows: 
Jan.   7,  an  invoice  of  $000;    Jan.  31,   $240;    Jan.  13, 
$400  ;  Jan.  20,  $320;  Jan.  28,  $1200;  I  jrive  a  note  at 
3  mon.  for  the  whole  amount:  when  is  it  dated? 
SOLUTION.  —  Start  at  the  date  of  the  1st  purchase,  Jan.  7;  then 
Any  date   will   do   to   Start      Purchases.  Disc.  Prod. 

with.  If  you  start  at  or  be-  0  0  0  X  0  = 
fore  the  1st  payment,  count  24  )  X  4=  9GO 
forward  for  the  multipliers  400X  6  =  2400 
and  quotient;  if  you  start  320  X  13=  41GO 
at  or  after  the  last  payment,  1200  X  2  1  =  25200 
count  backward.  If  you  start  97(50  ^32720^12 

between    the   first   and    last  ,0  ,          A     »       -        r      * 

JJence.  12  days  after  Jan.  I  =  Jan.  19, 

payments,  there  will  be  two     ,,  .  ,  .      .,, 

the  mean  time  or  date  of  the  note. 

columns  of  products,  one  of 

the  purchases  before  the  assumed  date,  and  the  other  of  those  after 
it;  in  that  case,  add  each  column,  take  the  difference  of  the  sums, 
and  divide  it  by  the  sum  total  of  debts:  this  quotient  must  be 
counted  forward  or  backward  from  the  assumed  date,  according  as 
the  products  after  or  before  that  day  preponderate. 

REMARK.  —  The  interest  gained  on  the  purchases  made  after  the 
mean  time,  should  be  just  equal  to  the  interest  lost  on  the  purchases 
made  before  it;  but,  if  the  quotient  which  fixes  the  mean  time  is 
not  an  exact  number  of  days,  the  two  interests  will  slightly  differ, 
according  to  the  size  of  the  fraction  neglected;  in  the  example 
above,  the  quotient  is  not  exactly  12  days,  but  llgg  days,  on  which 
account,  the  interest  before  Jan.  10,  will  be  a  little  more  than  that 
after  it,  but  as  a  part  of  a  day  is  not  recognized  in  dating  notes,  the 
answer  must  be  considered  correct. 

TO   AVERAGE   AN   ACCOUNT    HAVING   DEBITS   ONLY, 

RULE.  —  Start  at  the  earliest  day  a  debt  is  due,  and  multiply 
each  debt  by  the  number  of  days  (or  months)  its  dale  is  ajter 
the  day  fixed  on  ;  take  the  sum  of  these  products,  and  divide 
it  Ly  the  sum  total  of  the  debts  ;  the  quotient  will  be  the  number 

REVIEW.  —  341.  What  date  is  the  starting  point?  Would  any  other 
time  do  T  What,  if  you  start  at  or  before  the  first  payment?  What,  if  yon 
start  at  or  after  the  last  payment?  Whnt,  if  you  start  between  fie  tra* 
»nd  Insf  <Int«>fi  T  What  is  the  ml<»  for  averaging  ao  aceonnt? 


EQUATION   OF  PAYMENTS. 


291 


r-f  tlat/s  (or  months)  which,  added  to  the  day  first  fixed  on,  u-tll 
gice  Ike  mean  time  ryjnired. 

I'UOOF. —  Assume  the  day  the  last  debt  is  due  (or  ai.j 
other  day)  ;  determine  the  mean  time  again,  and  see  ii'  It 
agrees  with  the  one  already  obtained. 

ART.  342.  If  the  account  has  credits  as  well  as  debits, 
it  may  be  averaged,  on  the  same  principle,  by  this 

RULE. 

Slart  at  the  earliest  day  when  a  payment  is  made,  or  is  due 
in  cash;  counting  from  this  day,  form  the  product  for  each 
item,  both  on  the  Dr.  and  Or.  sides,  setting  each  product  in  a 
Dr.  or  Cr.  column,  according  to  the  item  multiplied.  Balance 
the  columns  of  products,  and  also  the  columns  of  items ;  the 
former  divided  by  the  latter  will  be  the  number  of  days  to 
be  added  to  the  earliest  date,  to  give  the  day  the  balance  of  items 
is  due. 

What  is  the  balance  in  this  account,  and  when  is  it  due  ? 
DR.  A  in  acc't  with  B.  CR. 


1855. 

Da.  Dol. 

•t 

Prod. 

1855. 

D.i. 

Dol.  ct. 

Prod. 

.Ian. 

H 

To  Invoice,  2m 

ia 

40() 

6^)0 

Mar 

1 

By  cash, 

0 

600 

0 

Kob. 

i 

»»        »      » 

81 

2tiO 

6^00 

Apr 

!•• 

'•  remittance 

Mar. 

10 

91             M          II 

70 

500 

3ft  (Oil 

Way  1, 

a 

250 

15250 

. 

•tf 

II         ><       » 

u 

800 

('.Sum 

May 

'20 

,.  cash. 

so 

376 

30000 

May 

16 

»         II       II 

187 

300 

41100 

Jun 

4 

,,  remittance 

2*H) 

fvwo 

June  25, 

lie, 

450 

522fK) 

1575 

974.50 

1675 

97460 

~G25)      68050' 

93  days  after  March  1  —  June  2. 

ins.  A  owes  B  $625,  due  June  2. 

SOLUTION. — Start  with  Mar.  1,  the  earliest  day  a  payment  is 
made  or  due.  Find  the  products,  both  on  the  Dr.  and  Cr.  side, 
using  the  dates  when  the  items  are  due  as  cash.  The  balance  of 
products,  68050,  divided  by  the  balance  of  items,  G25,  determines  the 
mean  time  to  be  93  days  after  Mar.  1  =  June  2. 

DEMONSTRATION. — In  order  to  be  due  Mar.  1,  the  Dr.  items 
must  suffer  a  discount  of  $155500  for  1  day,  and  the  Cr.  items  a 
discount  of  $97450  for  1  day.  The  Dr.  side  then  must  suffer  & 
discount  of  $58050  for  1  day,  more  than  the  Cr.  side.  Therefore, 
the  balance  due  Mar.  1  on  the  Dr.  side,  is  $025, /«*  the  discount  of 


REVIEW.— 341.  What  is  the  proof?  342.  What  is  the  rule  for  ave- 
raging an  account  that  has  credits  ?  Solve  the  example.  Demonstrate  ik« 
role. 


RAY'S   HIGHER   ARITHMETIC. 


$58050  for  1  day  ;  tint  the  disc:  tint  of  S5SOoO  for  1  day  is  (lie  snme 
as  the  discount  of  $G'J.j  for  93  days  (Art.  3-11);  and  $<>'2.\  du« 
March  1,  after  being  discounted  for  93  days,  is  the  same  as  $0'J5 
due  93  days  after  March  1  =  June  2. 

REMARK. — IP  the  balance  of  items  is  on  a  different  side  of  the 
licount  from  the  balance  of  products,  it  will  draw  interest,  instead 
if  suffering  discount,  for  the  time  determined  by  the  quotient;  or,  it 
till  lit  due,  (hat  long  be/ore  the  assumed  date. 

ART.  343.  The  rule  for  equation  of  payments  is  founded 
on  Bank  Discount,  and  assumes  that  the  interest  lost,  by 
paying  $100  five  days  before  due,  is  exactly  the  same  as 
the  interest  gained,  by  paying  $100  five  days  nfu-r  it  is 
due.  This  is  not  strictly  true;  for,  if  I  pay  $100,  5  days 
before  due,  my  exact  loss  is  not  5  days'  interest  on  $100, 
but  on  the  present  worth,  which  is  less  than  $100;  while, 
if  I  pay  $100,  5  days  after  due,  my  gain  is  exactly  5  days' 
interest  on  $100;  a  little  interest  is  thus  gained  by  the 
debtor,  or  rather  the  mean  time  of  payment  is  postponed, 
but  the  rule  being  confined  to  debts  of  short  date,  the 
error  is  small. 

EXAMPLES   FOR  PRACTICE. 

1.    What  is  the  mean  time  of  the  following  invoices: 
A  to  B.  DR. 


When 

due 

Dnys 

Products. 

ISM. 

utter 

M.iy 

15 

To  invoice  at  4  months. 

$800 

June 

1 

„                       <        „ 

71  in 

„ 

10 

,,                       4        „ 

'.too 

July 

^•0 

„         „            *        » 

(XX) 

Aug. 

1 

»         »            •        * 

fi<H) 

•» 

Ij 

ii         ii            *        » 

I'  HI) 

Sum  total, 

4000 

Ant.  Oct.  30,  1851. 
REMARK. — Start  with  Sept.  15,  the  day  the  first  debt  is  due. 

2.     DR.  E  in  acc't  current  with  F.  Ca. 


F«-h. 
Mar 

4 

•jo 

To  invoice  3  mon. 

.v,o 

—  IProd. 

May 

R 
"i; 

By  ca*h, 

l.-,o 
]•>  < 

~ 

Prod 

Apr. 

1 

.,        »      3     ,. 

,.-,(, 

- 

.Inn. 

H 

„  note,  I  iii'>n. 

:  14'  ' 

I 

» 

6 

.,        »     3    „ 

:K, 

-1 

,  Jul. 

' 

..       »     2    „ 

170 

- 

Am.  E  owes  F  $205,  due  Feb.  26. 

R  e  VIE  w. — 342.  ff  the  balance  of  items  is  on  a  different  side  of  the  ac- 
tountg  from  the  balance  of  products,  what  does  it  indicate?  On  what  sor 
»f  discount  i»  the  rmis  foundedT  To  whoso  advantage  is  this?  WhyT 


EQUATION   OF   PAYMENTS. 


293 


8. 


IT.  Wr  iff  fit  to  Mason  Jf  Giles. 


DR. 


1 

Dol. 

«t.  When  due 

Dav*  after 

Producu. 

18.V5. 

AprU  6. 

Feb. 

I 

Tc  invoice  3  months. 

900 

•JO 

»         ,     3        „ 

7«> 

Mar. 

10 

,     3        „ 

8<iO 

Apr. 

8 

„         ,  cash, 

fl'lO 

May 

10 

,     3       „ 

9nO 

.lun. 

15 

,,        ,     3       „ 

4no 

i       - 

Sum  total, 

4000 

.4ns.  Due  June  13. 

REMARK. — Start  with  April  8,  the  time  the  first  payment  is  due. 
4.     DR.  A  in  acc't  current  with  B.  CR. 


Mar:  l!l 

To  in  voici-  cash. 

;im 

^ 

Prod.  :  I-V1.. 

•in 

l?y  cnsh, 

4<iO 



Prod. 

AIT. 

•_'n 

,,        ,, 

S-X) 

— 

Mar. 

6 

„   remit.  Mar.  15. 

::un 

— 

.Inn. 

r> 

„        „ 

70;  ) 

_ 

.Inn. 

•jo 

.,   cash, 

•Jen 

_ 

May 

10    >.        »         „ 

i;m 

_ 

.lul. 

in 

»>      » 

frf*) 

— 

:  

Ans.  A  owes  B  a  balance  of  $1600,  due  April  23. 

REMARK. — Start  with  Feb.  20,  and  date  the  remittance,  Mar.  15, 
the  day  it  is  received. 

5.     DR.  C  in  acc't  current  with  D.  Ca. 


Jim. 

4 

To  in  voice  2  raon. 

•2--.ii 



Prod. 

Mar. 

K> 

Ily  cash. 

3.W|- 

I»-jd. 

K«V>. 

a 

,,        „     1    » 

1*1 

— 

•Jl 

»      •« 

»K)'_ 

lit 

4.-.D 

— 

A|.r. 

4 

„  nnteSmnn. 

8W»U 

AIPI-. 

2 

„         „       ca.-li, 

H«l 

— 

Slny  2" 

„  r.-init..  -Mny  25. 

r.'O:— 

Jun.  10 

„  Hrcep.  ItiiJu.  xi^ht.olKii— 

.  C  is  Cr.  $470,  due  Aug.  12. 

6.  I  owe  $012,  due  Oct.  16,  and  $500,  due  Dec.  20. 
If  I  pay  the  first,  Oct.  1,  15  daj*s  before  due,  when  should 
I  pay  the  last?  Ans.  Jan.  16,  next,  27  da.  al'tcr  due. 

7.  I  owe  $2150,  duo  Sept.  16;  I  pay  $500,  Aupr.  4: 
when  is  the  balance  due?  Ans.  Sept.  29. 

8.  I  owe  a  man  the  following  notes  :  one  of  $800,  duo 
May  1C;  one  of  $660,  due  July  1;  one  of  $940,  due  Sent. 
29:    he  wishes  to  exchange  them  for  two  notes  of  $1:200 
each,  and  wants  one  to  fall  due  June  1st;    when   should 
the  other  fall  due?  Am.  Sept.  !). 

0.    A    owes  $840,  due  Oct.  3 ;  ho  pays  $400,  July  1  ; 
$200,  Aug.  1  :  when  will  the  balance  be  due? 

Ans.  April  30,  next  ye. 

10.    I  owe  $3200,  Oct.  25;  I  pay  $400,  Sept.  15;  $800, 
Sept.  30:  when  will  the  balance  be  paid?     Ans.  Nov.  12 


294 


RAY'S    HIGHER   ARITHMETIC. 


11.  An   account  of  $-500   is  due  Sept.  10;  $500  arc 
paid  An-.  1;  $500,  Aug.  11;  §500,  Aug.  "21  :   when  will 
the  b.-ilunre  be  due?  Ans.  Nov.  9. 

12.  Exchange  the  five  following  notes  for  six  others,  each 
for  the  same  amount,  and  payable  at  equal  intervals.     Uno 
of  SI  200,  due  in  41  days;  one  of  §1500,  due  in  72  days; 
one  of  $2050,  due  in  80  days;  one  of  $1320,  due  in  110 
days;  one  of  $1730,  due  in  125  days;   total,  $7800. 

Ans  The  notes  are  $1300  each,  and  run  25,  50,  75, 
100,  125,  150  days  respectively. 

R  K  ii  A  R  K. — Since  the  notes  of  $1300  are  due  at  1,  2,  3,  4,  6,  6  in- 
tervals respectively,  use  these  numbers  as  multipliers :  the  quotient, 
25  diiys,  is  the  length  of  the  interval. 

13.  I   buy  property  for  $12000,-   -g   in    cash,  and  the 
balance  in   2  equal   payments,  at  3  and   6  mon. ;   I   pay  | 
down,  and  the  balance  in  3  equal  payments,  at  equal  in- 
tervals: what  is  the  interval?  Ans.  2  inon. 

14.  Exchange  3  notes,  $300  due  in  10  da.,  $500  due  in 
25  da.,  $1000  due  in  40da.,  for  $000  cash,  and  2  notes  for 
$550  and  $650  due  at  equal  intervals;  find  the  interval. 

Ans.  30  da. 

15.  Account  sales  of  lard  oil,  per  steamboat  Madison,  for 
acc't  of  X  and  Y,  Cincinnati. 


law. 

Dol.U. 

D.ir«  utlcr 
ilar.  9. 

Product!. 

Mar. 

9 

1  bbl.  lard  oil,  30  eal.  at  80  c  —30  days. 

31  20 

10     1     „      »      >t     39    „      ,    8"c  —  „      „ 

31  20 

12     8    „      „      „   318    „     ,    8i>  c  —  60      „ 

2.M  4(1 

15  10     „      ,        „    3H9    „      ,   78  o—  „      „ 

311  -ti 

19     1     „      „      „      40i  „      ,    80  c  —  30      „ 

3-Mii 

21   Iti     „      „      „    638    „     ,    75  c  —  60      „ 

478  5lJ 

22    3    „      „      ,,1  1«1  „     „  78  o  —  „      „ 

W21 

23  10    „     „     „    402   „     „  77  c  —  „      „ 

31W 

54 

50  bbl.    —      1995  Due  aa  cash,  May  15J18. 

1641 

67 

CHARfiKS. 

To  freight,  $15,   drayage, 

(3  •&  4825 

,,    pt/>ra£«,  labor,  coop'age   8| 

1 

„    fire  insiir.  ami  adv  535 

„    coHniiiwion   A  guaran- 

tee, 4  per  cent  

o™ 

N«"t  proceeds,  ilup  May  ICjio. 

123  -* 
111840 

Errors  and  oinisiiiou^  ^-xrcjited. 

EVANS  A  PRICK, 

New  Orli-HTiS. 

RE  MARK. — The  object  in  averaging  an  account  sales,  is  that  the 
consignor  may  draw  a  bill  of  exchange  for  the  net  proceeds,  to  fall 
*ue  on  the  day  of  equation,  without  loss  to  either  party. 


CO?.:iOUND   INTEREST. 


295 


10.    Account  sales  of  1000  boxes  star  candles,  for  ao- 
couut  of  Messrs.  A  and  B,  of  Cincinnati. 


Day-,  after  Products.] 

1854. 

I)  »1. 

at. 

Aj;r.  21. 

Apr.   21 

Sold    5  hxs.    150  Ib.  224  c.  cash 

33 

75 

May    11 

1  box,     30  Ib.  22i  c.      , 

6 

75 

Jun 

us 

1    „        30  Ib.  25  c    "   , 

7 

60 

Jul. 

13 

1    „        30  Ib.  il  c         , 

6 

90 

ft 

11 

10  bx.i.    300  Ih.  23  c    2  per  cent.  Off, 

67 

62 

Se>t. 

1 

500    „  16000  Ib.  23  c    at  fi 

mon. 

34.50 

IK) 

• 

2 

482    „  14460  Ib.  '£ic   at  6  mon. 

3325 

80 

1000        Due  as  ca*b,  Feb.  25 

28,  1855, 

G898 

• 

CHARGES. 

Apr. 

6 

To  freight  on  1000  boxes  .    210  23 

,,  drayHge,  storage,  ins.     104  64 

„  interest   oo    freight   to 

Feb.  28,  1855,  . 

,,  com.  and  nuttrantee  5 

' 

per  el.  on  f'.vi.v.  •:•_'. 

677 

39 

Net  proceeds  due  Keh.  25 

28,  1855,  «2iO 

93 

Errors  and  omissitms  exc«ptwl 

llK.NKV 

W'HITB, 

ART.  344.  Accounts  are  averaged  by  tables  of  interest, 
as  follows  : 

Tii Ice  out  (he  interest  at  any  rate,  on  each  item  of  (he  ac- 
count, counting  from  the  last  day  of  the  previous  year,  or  any 
other  convenient  tiny :  find  the  balance  of  interest,  and  refer 
to  the  table,  under  the  head  of  the.  same  rate,  for  the  time 
corresponding  to  this  balance:  this  will  be  the  average  time 
required. 

R  E  H  A  R  K. — Start  at  the  last  day  of  the  previous  year,  because  the 
date  of  each  item  will  show  how  many  months'  and  days'  interest 
must  be  taken. 


XXIX.  COMPOUND  INTEREST. 

ART.  345.  Compound  Interest  differs  from  Simple  In- 
terest in  allowing  the  Interest,  as  it  accrues  to  be  con- 
verted into  principal  ;  thus  producing  interes*  or.  interest 
and  increasing  the  amount  due,  much  more  kapidly  than 
simple  interest,  for  the  same  time  and  rate. 


R  K  v  i  E  w. — 345.  Does  compound  interest  differ  t.om  simple  interest! 


29G  HAY'S   HIGHER   ARITHMETIC. 

ART.  346.  Compound,  like  Simple  Interest,  maybe  pay- 
able  annually,  semi-annually,  quarterly,  &c. ;  but  the  rate 
per  cent,  is  <:iven  for  a  year,  it  bein<;  understood,  as  in 
simple  interest,  that  the  rate  semi-annually  is  one.  h«IJ\ 
and  the  rate  quarterly  is  one  quarter,  of  the  annual  rate. 

Strictly  speaking,  the  quarterly  and  semi-annual  rates  in  com- 
pound interest  are  not  one-fourth  and  one-half  of  the  annual  rate; 
for,  the  compound  interest  of 

$1  for  1  year,  at  8  %  annually,  is  .08  of  a  $. 

„     „         „        „  4  %  Bemi-annually,  is  .0816      „ 
.,     „         „        „  2  %  quarterly,  is  .08243216      „ 

ART.  347.    Compound,  like  Simple  Interest,  has  5  cases. 

CASE  I. 

Given,  the  principal,  time  and  rate,  to  find  the  compound 
amount  and  interest. 

RULE. — Find  the  amount  of  the  principal  for  1  interval,  at  the 
rate  for  that  interval;  then,  Hie  amount  of  (his  amount  for  an- 
other interval,  and  so  on  through  the  whole  number  of  intervals; 
if  there  be,  besides,  a  part  of  an  interval,  Jind  the  amount  of 
the  last  amount  for  this  time  by  the  rule  for  simple  interest. 
This  will  be  the  compound  amount  required,  and  the  compound 
interest  can  be  found  by  deducting  the  principal  from  it. 

NOTE. — By  interval  is  meant  a  year,  half-year,  or  quarter,  as  the 
interest  may  be  payable. 

Find  the  compound  amount  and  interest  of  $4600  for 
2yr.  3  mon.  15  da.,  at  6%,  payable  semi-annually. 

SOLUTION. — The  interval  is  6  months,  and  the  corresponding  rate 
is  3  <&.  The  number  of  intervals  being  4,  find  the  amounts  of  4  suc- 
cossive  principals  at  Sfe,  and  then  the  amoun*  of  the  last  amount, 
($5177.34),  for  3 mon.  15da.,  at  6^  per  annum:  the  result  ($5207.04) 
U  the  compound  amount;  deducting  the  principal,  $4600,  the  remain- 
dee,  $067.91,  is  the  compound  interest. 

REVIEW.— 346.  How  is  compound  interest  payable?  What  is  said  of 
the  rate  in  compound  interest?  Show  that  rate  quarterly  and  somi- 
annually  aro  not  i  and  J  of  the  yearly  rate  in  compound  interest?  317 
What  is  the  rule  for  finding  the  comp.  amount  and  interest  of  any  sum  for 
a  given  time  and  rate  ?  What  is  meant  by  interval  f 


COMPOUND  INTEREST.  297 

1.  Find  the  compound  amount,  and  interest,  of  $3850, 
for  4 yr.  7  mon.  16  da.,  at  5%,  payable  annually. 

AM.  $4826.59  and  $976.59 

2.  The  compound  interest  of  $13062.50,  for  1  yr.  10 
mon.  3  2  da.,  ato  %,  payable  quarterly.     Ans.  $2082.25 

3.  The  compound  amount  of  $1000  for  3  yr.  at  10  %, 
yayable  semi-annually.  Ans.  $1340.10 

4.  Find  the  difference  between  the  simple  and  comp.  int. 
of  $6320  for  5  yr.  8 mon.  6  da.,  at  6 % .     Ans.  $329 . 23 

5.  "What  sum,  in  2  yr.  5  mon.  27  da.,  at  6  %  simple  int., 
will  amount  to  the  same  as  $2000  at  comp.  int.,  for  the  same 
time  and  rate,  payable  semi-annually?      Ans.  $2016.03 

6.  Find  the  compound  int  of  $1000000  for  10  yr.  at 
,  payable  annually.  Ans.  $1593742.46 

7.  I  have  80  shares  .of  stock  ($50),  which  pay  a  divi- 
dend every  6  mon.  of  5  %  in  stock :  how  many  shares  will 
I  have  in  3  yr.  6  mon.  ? 

Ans.  112  shares  and  $28.40  of  another. 

8.  If  I  start  with  $5000,  and  increase  my  capital  15  % 
every  year,  what  will  it  be  in  6  years? 

Ans.  $11565.30 

9.  Find  the  comp.  int.  of  $1200  invested  at  8^  for  2yr., 
and  then  at  10  %  for  3yr.  Ans.  $662.97 

ART.  348.  As  the  operations  in  comp.  interest  arc  very 
laborious  when  the  time  is  long,  tables  are  used  to  facilitate 
the  work. 

To  find  the  compound  amount  and  interest  of  any  prin- 
cipal, for  any  time  and  rate  in  the  limits  of  the  table, 

RULE. 

Observe  at  what  intervals  the  interest  is  payable,  also  t7ie  num- 
ber of  such  intervals,  and  the  rate  corresponding  to  each.  Find 
from  the  table  the  compound  amount  of  $1  for  this  rale  and 
number  of  intervals,  and  multiply  it  by  the  given  principal. 

If  there  is  any  remaining  time,  calculate  the  amount  on  this 
product  for  that  time,  at  the  given  rate :  the  result  will  be  the 
compound  amount  for  the  whole  lime,  and  the  compound  intercfl 
•Mn  be  found  by  deducting  the  principal  from  it. 

RRTIEW. — 348.  What  are  used  to  facilitate  computations  in  compound 
Interest?  How  do  you  find  the  compound  amount  and  interest  of  any 
principal  for  any  time  and  rate  within  the  limits  of  the  table  ? 


298 


RAY'S   HIGHER   ARITHMETIC. 


Amount  of  $1  at  Compound  Interest  in  any  number  of  yean. 


lYr, 

2  per  cent. 

2>j  per  cent. 

3  per  cent. 

3^i  per  cent. 

4  per  cent. 

4>i  percent. 

1 

2 

a 

4 

6 

1.02110  IHKM) 
l.Otlll  0000 
l.C   12  08IX) 
1.0821  3210 
1.1040  8080 

1.0250  0000  1.0300  0000 
1.050U   2500  1.0009  OOOO 
1.0708  9002  1.0927  2700 
1.1038   128J1  1.1255  0881 
1.1314  0821  1.1592  7407 

1.0350  0000  1.0400  0000 
1.0712  2500  1.0810  (XKK 
1.1087  1  787  1  1.1248  64<)0 
1.1475  2300  1.1098  6850 
1.1870  8031  1.2100  6290 

1.0450  0000' 
1.0920  25001 
1.1411   6012 
1.1925  18,,0 
1.2401  8194 

6 

7 
8 
9 
10 

1.1201  6242 
I.148o  85,i7 
1-1710  59:!8 
1.1950  9257 
1.218!)  9442 

1.1590  9342 
l.lSSii  8070 
1.2184  02  JO 
1.2488  0297 
1.2800  8454 

1.1940  6230 
1.2298  7387 
1.2ii>7  7008 
1.3047  7318 
1.3439  1038 

1.2292  5533 
1.2722  7920 
1.3108  0904 
1.3028  9735 
1.4105  9876 

1.2053  1902 
1.3159  3178 
1.3085  0900 
1.42:i3  1181 
1.4S02  4428 

1.3022  6012 
1-3008  G183 
1.4221  0001 
1.4800  9514 
1.5529  6942 

11 
12 
l.i 
It 
15 

1.2433  7431 
1.2  .,82  4179 
1.2J3.J  0003 
1.3191  7870 
1.3408  0834 

1-3120  8000 
1.  34-18  8882 
1-3785   1104 
1.4129  7382 
1.44H2  'J8i7 

1.3842  3387 
1  4207  OUSJ 
1.4.j85  3371 
1.5125  8972 
1.5579  0742 

1.4599  0972 
1.6110  08  JO 
1.5039  6uOii 
1.0180  9452 
1.6753  4883 

1.5394  5406 
1.C010  3222 
1.6050  7351 
1.7313  7045 
1.8009  4351 

1.6228  5305 
1.6908  8143 
1.7721  9610 
1.8019  4492 
1.9352  8244 

1(5 
17 
18 
19 

20 

1.3727  8570  1-4815  05fi2 
1.40112  4112  '  1.621J  182; 
1.4282  4025  l.509«  6872 

1-4018  1117  i.598o  tola 

1.4809  4740  1.03SJ  1044 

1.C047  0044 
1.0528  47  03 
1.7024  330J 
1.75.J5  0005 
1.80U1  1123 

1.7339  8004 
1.7J40   7000 
1.8574  8U20 
1.9225  0132 
1.9897  888  J 

1.8729  8125 
1.9479  0050 
2.0258  1052 
2.1008  4918 
2.1911  2314 

2.0223  7015 
2.1133  7681 
2.2084  7877 
2.3078  6031 
2.4117  1402 

21 
22 
•2.1 
21 
25 

1.515C  CG34 
1-5409  7J07 
1  .5708  9920 
1-0084  3725 
1.040J  0099 

1.6795  8185  1.8fl02  9457 
1.7215  7140  1.  'Jl.il  0341 
1.7040  10J8  1.1)735  8051 
1.8H«7  2095  2.0327  U411 
1  8539  4410  2.0UJ7  7793 

2.0504  3147 
2.1315   1158 
2.20,;l   1448 
2.2833  2849 
2.3032  4498 

2.2787  6807 
2.3099  1879 
2.4647  1550 
2.5.J33  0417 
2.6058  3033 

2.5202  4116 
2.6330  6201 
2.7521  6035 
2.87uO  1383 
3.0054  3446 

20 
27 
28 
29 
30 

1.0734  1811 
1.70J8  8048 
1.7410  2421 
1.7758  44ii9 
1.8113  G158 

1.9002  9270  2.1505  9127 
1.9478  0002;2.2-.n2  8901 
l.9Uo4  9502  2.2879  27iJ8 
2.04(4  073U  12.3565  0501 
2.0975  6;58  2.4272  0247 

2.4459  6850 
2.5315  0711 
2.0201  719J 
J.7118  7798 
2.8007  9370 

2.7724  C979 
2.8833  0858 
2.U9S7  0332 
3.1180  6145 
3.2433  9751 

3.1400  7901 
3-2820  0950 
3.429ii  9999 
3.6840  3049 
3.74o3  1813 

31 

32 
33 
34 
35 

1-8475  8882  2.1500  0077 
1-8845  4059  2.2037  6094 
l.!)222  JU40  2.2588  6080 
l.UOOii  7UI3  2.3153  2213 
1.9998  8955  2.3732  0519 

2.5000  8035 
2.5750  8270 
2.0523  3024 
2.7319  0530 
2.8138  0245 

2.9050  3148 
3.00,i7  0759 
3.1119  4235 
3.2208  0033 
3.3335  9045 

3.3731  3341 

3.5080  5875 
3.0483  8110 
3.7943  1034 
3.9400  8899 

3.9138  5745 
4-0899  8104 
4.2740  3018 
4.4003  01.54 
4.0073  4781  , 

36 
37 
38 
39 
40 

2.0398  8734 
2.080<i  8509 
2.1222  9879 
2.1647  4477 
2.2080  3900 

2.4325  3532 
2.4933  4870 
2.555(5  8242 
2.0195  7448 
2.0850  C384 

2-8982  7833 
2.9852  2068 
3.0747  8348 
3.1070  2i>98 
3.2020  3779 

3.4502  6611 
3.5710  2543 
3.091)0  1132 
3.8253  7171 
3.9092  6972 

4.1039  3255 
4.2080  8980 
4.4388  1345 
4.6103  0599 
4.8010  2003 

4.8773  7846  ; 
5.0908  0049  , 
5.3202  1921 
6.5058  0908 
5.8103  0454 

41 
42 
43 
44 
45 

2.2522  0046 
2.2972  4447 
2-3431  8930 
2.3900  5314 
2.4378  6421 

2-7521  9013 
2.8209  9520 
2.8915  2008 
2.  9038  0808 
3.0379  0328 

3.3598  9893 
3.4006  £589 
3.5045  1077 
3.6714  6227 
3.7815  9584 

4.0978  X381 
4.  it  12  5799 
4.3897  0202 
4.5433  4100 
4.7023  6855 

4.9930  6145 
5.1927  8391 
5.4004  9527 
5.GIG5  1508 
5.8411  7568 

£.0781  0094 
6.3510  1548 
6.6374  3818 
6.9301  2290 
7.2482  48431 

46 
47 

48 

i  49 
50 

2.4808  1129 
2.5303  4351 
2.5870  70.19 
2.0388  1179 
2.0915  8803 

3.1138  f>08fi 
3.1910  9713 
3.2714  8950 
3.3532  7080 
3.4371  0872 

3.8950  4372 
4.0118  9503 
4.1322  5188 
4.25,;2  1944 
4.3839  OJO-2 

4  8009  4110 
5.0372  8404 
.2135  8898 
.39!  10   C4.VJ 
.5849   2080 

6.0748  2271 
0.3178  1502 
o.5700  2S21 
ti.8333  4937 
7.10oO  KKX> 

7.5744  1061 
7.9152  6819 
8.2714  5557  ; 
8-0430  7107 
9.0320  3027  ' 

51 

.12 
53 
M 

,v, 

2.7454  1979 
2.8003  2819 
2.8503  3475 
2.  HI  34  0144 
2.9717  3067 

3.5230  3044 
3.0111   12)S5 
3.7013  9010 
3.7939   2491 
3.8887  7303 

4.5154  2320 
4.0508  K.V.HI 
4.7904   1247 
4.9341  2485 
5.0821  4859 

.7801  9030 
.9827  1327 
0.1921  0824 
0.4088  3202 
6.6331  4114 

7.3900  50-.8 
7.G8:.5  8871 

7.9940  522n 

8.3138  1435 
8.6403  6692 

9.4391  0490 
9.8038  6403 
10.3077  3853 
10.7715  8077 
11.2503  0817 

COMPOUND   INTEREST. 


299 


Amount  of  $1  at  Compound  Interest  in  any  number  of  years. 


Yr 

5  per  ce>t. 

C  per  cent. 

7  per  cent. 

8  per  cent. 

9  per  cent 

10  per  cent. 

1 

5 

4 
6 

1.0500  (MM 

1.1025  ooi 

1.1571.  250 
1.2155  0..3 
1.2702  blO 

1.0000  (MX) 
1.1  230  000 
1.1910  100 

1.2  121   77( 
1.3382  250 

1.0700  OOf 
1.1449   000 
1.2250  431 
1.3107  9iiO 
1.4025  517 

1.0800  (MX) 
1.1004  000 
1.2597  120 
1.3001  890 
1.4093  281 

1.0900  OIK 
l.lSSl   (KM 
1.2950  290 
1.4115  811 
1.538J  240 

i.iooo  aw 

1.2100  WM) 
1.3310  0"0 
1.4,41  000 
1.0103   100 

C 
7 
8 
9 
10 

1.3400  Oil! 
1.4071  (K)4 
1.4774  654 
1.5513  2rfJ 
1.0288  !HJ 

1.4185  191 
1.5030  303 
1'6933  481 
1.0894  790 
1.71)08  477 

1.5007  304 
1.UI57  815 
1.7181  8,i2 
1.8381  5'J2 
1.9071  514 

1.58G8  74.'! 
1.7138  243 
1.8509  302 
1-9990  04:i 
2.1589  250 

1.C771  001 
1.8280  391 
1.9925  G2o 
2.1718  933 
2.3073  037 

1.7715  CIO 
1  9487  171 
2.1435  8S8 
2.3579  477   1 
2.5937  423   j 

11 

8 

u 
15 

1.7101  3!»4 

1.7958  6.,3 
1.8.S5.;  4'.M 

1.9799  ;u.i 

2.  0789  282 

1.8982  98G 
2.0121  Oo5 
2.1329  283 
2.2H09  040 
2.3900  582 

2.1048  520 
2.2521  910 
2.4098  450 
2.5785  342 
2.7590  315 

2.3T.16  3!H) 
2.5181  701 
2.7190  237 
2.9371  93  i 
3.1721  C91 

2.6804  264 
2.812J  048 
3.0058  04  o 
3.3417  270 
3.0421  825 

2.8531  167 
3.l:;s4  284 
3.4522  712 
3.7974  983 
4.1772  482 

Ifi 

17 

18 

in 

20 

2.1828  7Ki 
H-2920   183 
'£  4000  l'J2 
2.5209  502 
2.6532  1)77 

2.54OT  517 
2.0927   728 
2.8543  3!)2 
3.0255  9'.l5 
3.2071  350 

2.9521    038 
3.1588   152 
3.379!)  323 
3.r,105  2,V» 
3.8li'JO  t>45 

3.4259  42(i 
3.7(KH>  181 
3.9!)i;0  195 
4.3157  "11 
4.0009  571 

3.9703  059 
4.3270  334 
4.7171  201 
5.1410  013 
6.WM4  108 

4.5949  730 
5.0544  703 
5.6599  173 
6.1159  090 
6.7275  000 

21 
22 

21 

2> 
2.-. 

2.7859  02li 
2-9252  lil'7 
3.0715  2.18 
3.2250  999 
3.3803  549 

3.3995  630 
.'•..Go:i5  374 
3.8197  4'.)7 
4.  ol  S.I  340 
4.2U18  707 

4.140.-.  C21 
4-4301  017 
4.7405  2:ii) 
5.072:1  07o 
5.4274  320 

5.0338  -337 
5.4305  404 
6.8714  CJ7 
0.3411  807 
0.8484  752 

0.1088  077 
C.0580  OO4 
7.2578  745 
7.9110  fc-'!2 
8.0230  fc07 

7.4002  499 
8.1402  749 
8.9543  024 
9.8497  327 
10.8347  059 

2: 
27 
28 
29 
30 

3.5550  727 

3.7334  5ii3 
3.H20I   291 
4.lli;i   :;5ii 
4.321'.)  424 

4-5493  830 
4.822.1  4;V.) 
5.  11  If,  867 
5.4183  879 
5.7434  912 

5.8073  52!) 
11.2138  070 
O.W88  384 
7.1142  571 
7.0122  650 

7.3903  .132 
7.9880  (.15 
8.0271  OU4 
9.3172  74!) 
10.002U  609 

9.3991  679 
M.2450  821 
11.1071   395 
12.1721    821 
13.207U  7»5 

11.9181   705 
13.1099  942 
14.4209  93K 
15.  Si.30  930 
17.4494  023 

3i 

:!2 
33 
34 
35 

4.5380  3!I5 
4.7ii4!>  41;': 
5.0031  8*5 
5.2533  480 
5.5100  154 

<5.<1881   (HKi 
0.4,');»  81  >7 
(i.8405  Slf-l 
7.2510  2."-J 
7.0800  808 

8.1451   129 

8.7152  708 
9.3253  3!)S 
9.9781  i:5.r> 
10.0705  815 

10.8076  094 
11.7370  830 
12.0700  490 
13.0901   33ii 
14.7853  443 

14.4617  695 
15.7t>13  288 
17.1820  2o4 
18.7284  loy 
20.4139  079 

19.1943  425 
21.1137  708 
23.  2251    544 
25.5476  099 
28.1024  369 

3f, 
•XI 
38 

at 

40 

5.7918  101 
0.0814  <Hi!» 
6.3854  773 
(i.7047  512 
7-0399  887 

8.1472  520 
8.03IJO  871 
9.1542  524 
9.70:55  075 
10.2857   179 

11.4239  422 
12.22:tO   181 
13.0792  7W 
13.9948  214 
14.9744  67f 

15.9081  718 
17.2450  256 
18.0252  750 
20.1152  977 
•i  1.  7245  215 

22.2512  250 
24.2538  353 
20.4300  805 
28.8159  817 
31.4094  200 

30.9126  805 
34.0039   486 
37.4043   434 
41.1447  778 
45.2592  656 

•II 
42 
4:t 
44 
43 

7.3919  882 
7.7015  87  (J 
8-1  49(5  00!t 
8.5571  603 
8.9850  078 

10.9028  filO 
11.5570  327 
12.2504  640 
12.9854  819 
13.7040  108 

10.0220  C99 
17.1442  508 
18.3443  648 
111.  0284  51)0 
21.0024  518 

23.4624  832 
25.  3394  81!) 
27.3006  404 
29.5559  717 
31.9204  494 

34.2362  67!) 
37.3175  320 
40.0701  098 
44.3309  597 
48.3272  801 

49.7851   811 
54.7030  992 
00.2400  G92 
66.21*)  701 
72.8901  837 

40 
47 
48 
4!) 
50 

9.4342  582 
9.9059  711 
10.4012  097 
10.9213  331 
11.4073  U98 

14.590t  875 
15.4059  107 
10.39:18  717 
17.3775  040 
18.4201  543 

22.4726  234 
24.1U57  070 
25.728!)  (Mi5 
27.5299  300 
29.4570  251 

31.4740  853 
37.2320  122 
40.2105  731 
43.4274  190 
40.9010  125 

52.6767  419 
57.4176  48i, 
62.5852  370 
08.2179  083 
74.3575  201 

80.1795  321 
88.1!174  85=1 
97.0172  338 
06.718!)  672 
17.3908  529 

51 
52 
53 
54 
55 

12.0407  698 
12.0428  083 
13.274'J  487 
13.9381}  901 
14.6350  309 

19.5253  035 
20.0908  853 
21  9:i8li  985 
23.2550  20-1 
24.<>50:i  21  « 

31.5190  168 
33.7253  480 
30.08f>l  224 
38.0121   509 
41.3150  015 

50.6537  415 
54.7000  408 
59.0825  241 
63.8091  2W 
68.9138  561 

81  .0490  969 
88.3441  69 
90.2951  449 
104.9017  079 
114.4082  61  ii 

129.1299  382 
112.0129  320 
50.2472  252 
1?1.  8719  477 
189.0591   425 

300  RAY'S  HIGHER  ARITHMETIC. 

10.  Find  the  compound  amount  of  $750,  for  I7yr.  at 

6  %•>  payable  annually.  AM.    §2019.58 

11.  Of  §5428,  for  33  yr.,  at  5  %  annually.     • 

An*.  $27157.31 

12.  The  compound  interest  of  $1800  for  14  yr.,  at  S%, 
payable  scmi-annually.  Am.  §3597.07 

13.  If  §1000  is  deposited  for  a  child,  at  birth,  and  draws 

7  %   comp.  int.,  payable  semi-annually,  till    it    is  of  ago 
(21  yr.),  what  will  it  amount  to?  Ans.  §4241.20 

14.  Find  the  compound  amount  and  int.  of  §9401.50 
for  19  yr.  4  mon.,  at  9  %,  payable  semi-annunlly. 

Ant.  Ain't.  §51576.68,  Int.  §42175.18 

15.  The  comp.  int.  of  §1176.  80  for  10  yr.  lOmon.  10  da., 
at  10  %,  payable  quarterly.  Ans.  §2263.75 

16.  The  comp.  amt.  of  §5000  for  12  yr.  5inon.  22  da,, 
at  12%,  payable  semi-annually.  Ans.  §21405.37 

17.  The  comp.  int.  of  §8025  for  18  yr.  7  mon.  9  da.,  at 
8%,  payable  semi-annually.  .AHS.  §26523.27 


If  the  time  is  beyond  the  limits  of  the  table,  use  this 

RULE. 

Separate  the  given  time  into  two  or  more  period*,  each  within 
the  limit*  of  the  table.  Find  the  compound  amount  of  lite  princi- 
pal for  the  first  of  these  periods,  then  the  compound  amount  of 
this  amount  for  the  second  period,  and  so  on.  The  last  amount 
will  be  the  compound  amount  required,  from  which  the  compound 
interest  can  be  found  as  before. 

18.  Find  the  compound  amount  of  $1  000  for  1  00  yr.,  at 
10  %•>  payable  annually.  Ans.  §13780612.34 

10.  The  comp.  amt.  and  int.  of  §3600  for  15yr.,at  S%, 
payable  quarterly.  Ans.  Amt.  §11811.71,  Int.  §8211.71 

20.  The  comp.  int.  of  $4000  for  40  yr.  at  5  %,  payable 
Bcmi-annually.  AHS.  §24838.27 

21.  The  comp.  amt.  of  §2500  for  32  yr.  8  mon.  6  da.,  at 
7  %i  payable  scrai-annually.  Ans.  §23691.95 

22.  Find  the  comp.  int.  of  §1200  for  27  yr.   11  ir.on. 
4  da.,  at  12  %,  payable  quarterly.         AHS.  §31404.74 

ART.  349.  CASE  II.  —  Given,  the  principal,  time,  and 
interest  or  amount,  to  find  the  rate. 

R  E  v  i  E  w.—  348.  What,  if  the  time  is  beyond  the  limits  of  the  tnl.'«  ? 


COMPOUND  INTEREST.  301 

RULE. — If  the  interest  be  given,  add  it  to  the  principal  to  gel 
the  amount;  then  divide  Ike  amount  by  the  principal;  the  quotient 
will  be  the  amount  of  §1  for  the  given  time  and  rate.  Look  in  the 
table  opposite  the  given  number  of  intervals  until  this  amount 
ii  found;  the  rate  per  cent,  at  the  head  of  the  column  will  be 
the  one  required. 

NOTE. — If  the  time  contains  a  part  of  an  interval,  find  the 
amounts,  for  that  time,  of  the  numbers  in  the  table,  before  com- 
paring them  with  the  quotient. 

AT   WHAT   RATE,    BY   COMPOUND  INTEREST, 

1.  •\Vill$1000amountto$1593.85in8yr.?  Ans.G%. 

2.  $3600  yield  $0332.51  int.  in  15yr.?     Ans.  *l  %. 

3.  $13200  amt.  to  $48049.58,  in  26yr.  5  mon.  21  da.? 

A  us.  5  % . 

4.  $2813.50  amt.   to  $13276.03,   in    17  yr.    1  mon. 
It  da.,  int.  payable  scmi-annually  ?  Ans.  9%. 

SPOOESTION. — The  rate  found  by  the  rule  in  this  example  is  the 
Bemi-anuual  rate,  and  must  be  doubled  to  get  the  rate  per  annum. 

5.  $7052.18  yield  $17198.67  int.,  payable  quarterly, 
in  11  yr.  Union.  3  da.?  Ans.  10  %. 

G.  At  wl^t  rate  will  any  sum  double  itself  by  compound 
int.  in  8,  10,  12,  15,  20  yr.? 

Ans.  1st,  little  over  9$;  2d,  between  7  and  8$:  3d, 
(not  quite  6  %  ;  4th,  not  quite  5%  ;  5th,  little  over  3i  %. 

ART.  350.  CASE  ITT. — Given,  the  compound  interest, 
the  time,  and  rate,  to  find  the  principal. 

RULE. — Assume  $1  for  tlie  principal;  determine  the  compound 
interest  on  this  supposition,  and  divide  the  yicen  compound  inte- 
rest by  it. 

PROOF. — With  the  principal  thus  found,  calculate  the  compound 
interest  for  the  given  time  and  rate ;  if  it  agrees  with  the  given 
compound  interest,  the  work  is  right 

NOTE. — To  get  the  amount,  add  the  principal  to  the  interest. 

1.  What  principal  will  yield  $52669.93  compound  int> 
in  25  yr.,  at  6  %  per  annum  ?  Ans.  $16000. 

REVIEW. — 319.  What  u  Case  2?  The  rule?  If  tho  time  contain  a 
part  of  an  interval  what  U  necessary  ?  360.  What  ia  Case  3  ?  The  rule 
The  proof? 


302  RAY'S   HIGHER   ARITHMETIC. 

2.  What  sum,  in  6yr.  2  mon.,  will  yield  $1625.75  com- 
pound int.  at  7%,  payable  semi-annually  ?    Ans.  $3075. 

3.  What  sura,  at  10  %,  payable  quarterly,  will  produce 
$3598.61  comp.  int.  in  3yr.  6  mon.  9  da.  ?     Ans.  $6640. 

4.  What  principal,  in  37 yr.,  at  5  %  per  annum,  will  pro- 
duce $0891.61  compound  interest?         Ans.  $1356.24 

5.  What  sum,  in  9-iyr.,  at  8$,  payable  semi-annually. 
will  yield  $31005.76  compound  int.  ?  AM.  $28012.63 

ART.  351.  CASE  IV. — Given,  the  compound  amount, 
the  time,  and  rate,  to  find  the  principal. 

RULE. — Assume  SI  for  the  principal;  determine  the  compound 
amount  on  this  supposition,  and  divide  the  given  compound  amount 
by  it. 

PROOF. — With  the  principal  thus  found,  calculate  the  compound 
amount  at  the  given  time  and  rate ;  if  it  agrees  with  the  givun 
compound  amount,  the  work  is  right 

NOTES. — 1.  To  get  the  interest,  subtract  the  principal  from  the 
amount. 

2.  If  the  compound  amount  is  a  debt  not  yet  due,  the  principal  is 
its  present  worth  by  compound  interest,  and  the  difference  between 
the  debt  and  present  worth  is  the  compound  discount  of  the  dell,  or 
the  comp.  int.  of  its  present  worth  for  the  given  time  and  rate. 

1.  What  principal   in  7  yr.,  at  4  %  compound   interest, 
will  amount  to  $27062.85?  AM.  $20565.54 

2.  Find  the  present  worth  of  $14625.70,  due  in  5  yr. 
9  mon.,  at  6  %  comp.  int.,  payable  semi-annually. 

AM.  $10409.77 

3.  The  discount   on   $8767.78,  due    in   12  yr.   8mon. 
25 da.,  at  5  %  comp.  int.  Am.  $4058.87 

4.  The  present  worth  and  comp.  discount  of  $48941.28 
due  in  10  yr.  4  mon.  6  da.,  at  10  %,  payable  quarterly. 

Am.  P.  W.  $17606.60,  C.  D.  $31334.68 
5    The  present  worth  of  $100000,  due  in  50 yr.,  at  9^ 
jomp.  int.  Ans.  $1344.85 

6.  The  difference  between  the  present  worth,  at  simpU 
and  at  comp.  int.,  of  $34058.75,  due  in  20  yr.,  at  6  %. 

Am.  $4861.57 

R  B  vi  *  w.  — 351.  What  is  Case  4?  The  rule?  The  proof?  Ilow  k 
the  comp.  int.  then  found  T 


ANNUITIES.  303 


ART.  352.  CASE  V.  —  Given,  the  principal,  rate,  and 
compound  interest  or  amount,  to  find  the  time. 

KULE.  —  If  the  interest  is  given,  add  it  to  the  principal  to 
get  the  amount ;  (hen  divide  the  amount  by  the  principal ;  ike 
quotient  will  be  the  compound  amount  of  §1  for  the  time  and  rate. 

Look  in  the  table  under  the  head  of  the  gicen  rate  until  tliit 
number  is  found;  the  corresponding  number  in  the  left  hand 
column  will  be  the  number  of  years  (or  intercah)  required. 

But  if  the  number  can  not  be  found  exactly  in  the  table,  take 
out  the  number  of  years  (or  intervals)  corresponding  to  the  smaller 
one  of  the  two  numbers  between  which  it  lies,  and  the  remainder 
of  the  lime  will  be  such  part  of  a  year  (or  interval)  as  the  differ- 
ence between  the  number  taken  and  the  quotient,  is  of  the  difference 
between  that  smaller  one  and  the  next  larger. 

The  part  of  a  year  (or  interval)  thus  determined  can  be  con- 
verted into  months  and  days,  and  annexed  to  the  whole  number  of 
years  (or  intervals)  already  taken  out. 

PROOF. — With  the  time  thus  found,  calculate  the  compound 
amount  of  the  given  principal  at  the  given  rate ;  if  it  agrees  with 
the  given  compound  amount,  the  work  is  right. 

1.  In  what  time  will  $8000  amount  to  $12000,  at  6  % 
compound  interest?  -4ns.  6  yr.  11  mon.  15  da. 

2.  In  what  time  will  any  sum  of  money  double  itself,  by 
compound  interest  at  5,  C,  7,  8,  10  %  ? 

AM.  14  yr.  2 mon.  13 da.;  11  yr.  10  mon.  21  da.;  lOyr. 
2  mon.  2t)  da.;  9  yr.  2  da. ;  7  yr.  3  mon.  5  da. 

3.  In  what  time  will  $5200  draw  $1308  comp.  int.  at 
6%  payable  scini-annually  ?       Ans.  3  yr.  9  mon.  16  da. 

4.  In  what  time  will  $12500  amount  to  $18000,  by 
comp.  int.  at  10  %,  payable  quarterly  ? 

Ans.  3yr.  8 mon.  9  da. 

5.  In  what  time  will  $9862.50  amount  to  $22570.15 
by  compound  interest  at  12  %,  payable  semi-annually  ? 

Ans.  7  yr.  1  mon.  7  da. 


XXX.   ANNUITIES. 

ART.  353.     The  principal  application  of  Compound  In- 
terest is  to  Annuities. 

REVIEW.— 352.  What  is  Ca«e  6?    Tberulo?    Theproof? 


304  RAY'S  HIGHER  ARITHMETIC. 

An  Annuity  is  an  estate  which  entitles  its  owner  to  the 
payment  of  a  fixed  sum,  at  regular  intervals  of  time. 

Annuities  comprise  Life-insurance,  Hents, Dowers, Leases, 
Life-estates,  Survivorships,  Pensions,  &c. 

It  is  called  Annuity,  from  the  Latin  annus,  meaning  a 
year,  because  the  interval  between  the  payments  is  generally 
a  year,  though  sometimes  it  is  half  or  quarter  of  a  year. 

A  Perpetuity  is  an  annuity  which  continues  forever. 

An  Annuity  certain  begins  and  ends  at  fixed  times. 

A  contingent  Annuity  begins  or  ends  with  the  happening 
of  an  uncertain  event;  as,  the  birth  or  death  of  one  jr 
more  persons. 

ART.  354.     An  immediate  annuity  begins  at  once. 

A  deferred  annuity,  or  annuity  in  reversion,  begins  at  a 
future  time. 

The  forborne  or  final  value  of  an  annuity,  is  the  sum  of 
the  compound  amounts  of  all  its  payments  at  an  assumed 
rate,  from  the  time  each  is  due,  to  the  end  of  the  annuity. 

The  present  value  of  an  annuity,  is  the  present  worth  of 
the  final  or  forborne  value;  that  is,  such  a  sum  as  put  out  at 
compound  interest  as  proposed,  will,  at  the  expiration  of 
the  annuity,  amount  to  its  final  value. 

The  value  of  a  deferred  annuity  at  the  time  it  commences, 
may  be  called  its  initial  value;  its  present  value,  is  the 
present  worth  of  its  initial  value,  at  an  assumed  rate  of 
compound  interest. 

An  annuity  is  said  to  be  worth  as  many  years1  purchase,  as  its 
present  value  contains  one  of  its  payments. 

An  annuity  begins,  not  at  the  time  the  first  payment  is  made,  but 
one  interval  bofore;  if  an  annuity  begins  now,  its  first  payment  will 
be  a  year,  half-year,  or  quarter  of  a  year  hence,  according  to  the  in- 
terval between  the  payments. 

CASE  I. 

ART.  355.  Given,  the  rate,  the  payment,  and  the  inter- 
val, to  find  the  initial  value  of  a  perpetuity. 

RKVIF.W. — 353.  What  are  annuities?  What  do  theycompri.se?  Why 
10  called  ?  What  is  a  perpetuity  ?  An  annuity  certain  ?  A  contingent  an- 
nuity? 35-1.  Wbnt  is  an  immediate  annuity  ?  A  deferred  annuity  ?  What 
is  the  forborne  or  final  value  of  an  annuity?  Its  present  value?  The  itii 
ual  value  of  a  deferred  annuity  ?  355.  What  U  Case  1  ? 


ANNUITIES.  305 


RULE.  —  Find,  by  Case  III  of  Simple  Interest,  the  princijml 
tohose  interest  at  the  given  rate,  for  the  given  interval,  equals  tht 
given  payment  :  this  will  be  the  initial  value  required. 

NOTE.  —  If  the  perpetuity  is  immediate,  its  initial  value  is  its 
present  value. 

What  is  the  initial  value  of  a  perpetual  lease  of  $250 
«  year,  allowing  G  %  interest? 

SOLUTION.  —  The  initial  value  must  &  \ 

be  Ihe  principal,  which,  at  G  %,  yields  fQQ 

$250   interest   every  year;    it  is   found  ~^.C)rn  A  nn  rt 

by  Art.  310.     The  operation  amounts  to  >u  °    "  °  u>u  u  u  u 


dividing  the  given  payment  by  the  interest  of      Ans.  $4160.  663 
SI  for  the  interval  and  at  the  rate  proposed. 

1.  What  is  the  initial  value  of  a  perpetual  leasehold  of 
$300  a  year,  allowing  6  %  interest?  Ans.  $5000. 

2.  What  must  I  pay  for  a  perpetual  lease  of  $756.40 
a  year,  to  secure  8  %  interest?  Ans.  $9455. 

3.  Ground  rents  on  perpetual  lease,  yield  an  income  of 
$15642.90  a  year:  what  is  the  present  value  of  the  estate, 
allowing  7  %  interest?  An*.  $223470. 

4.  What  is  the  initial  value  of  a  perpetual  leasehold  of 
$1600  a  year,  payable  seini-annually,  allowing  5  %    in- 
terest, payable  annually?  Ans.  $32400. 

SUGGESTION.  —  Here  the  yearly  payment  is  $1620,  by  allowing  5 
per  cent,  interest  on  the  half-yearly  payment  first  made. 

5.  What  is  the  initial  value  of  a  perpetual  leasehold  of 
$2500  a  year,  payable  quarterly,  interest  6  %  payable  semi- 
annually;  G%  payable  annually;  6  %  payable  quarterly? 

Am.  $41979.161;  $42604.  1.6i;  $41666.  66f 

CASE  II. 

ART.  356.  To  find  the  present  value  of  a  deferred  per- 
petuity, when  the  payment,  the  interval,  the  rate,  and  tho 
time  the  perpetuity  is  deferred,  are  known. 

RULE.  —  Find  the  initial  value  of  Ihe  perpetuity  by  the  last 
rule  ;  then  find  the  present  worth  of  this  sum  for  the  time  the 
perpetuity  is  deferred,  l>y  Case  IV  of  Compound  Interest;  this 
will  be  the  present  value  required. 


REVIEW.— 355.    What  is  the  rule?     If  tho  perpetuity  i»  immediate, 
what  is  the  initial  value?     Solve  tho  example. 
•215 


;j06  RAY'S  HIGITER  ARITHMETIC. 

Find  the  present  value  of  a  perpetuity  of  $250  a  year, 
deferred  8  yr.,  allowing  0  %  int. 

SOLUTION.  —  Initial  value  of  perpetuity  of  $250  a  year,  by  last 
rule  =  $41GtJ.(iGj.  The  present  value  of  $4166.66},  due  Syr.  hence, 
at6#,  comp.  int.,  =  f41GG.66|-M.593S481  (Art.  351).  Usethecon- 
tracted  method,  reserving  3  decimal  places  :  the  quotient,  $2014.22. 
|»  the  present  value  of  the  perpetuity. 

1.  Find  the  present  value  of  a  perpetuity  of  $"780  a  yr., 
ti;  commence  in  12  yr.,  int.  5  c/0.  Ans.  $8680.  G6 

2.  Of  a  perpetual  lease  of  $160  a  year,  to  commence  in 
3  yr.  4  inon.,  int.  7  %.  Ans.  $1823.28 

3.  Of  the  reversion  of  a   perpetuity  of  $540  a  year, 
deferred  lOyr.,  int.  6  %.  Am.  $5025.55 

4.  Of  an  estate  which,  in  5  yr.  is  to  pay  $325  a  yr.  for 
ever;  int.  8  %,  payable  semi-annually.     Ans.  $2600.67 

5.  Of  a  perpetuity  of  $1000  a  year,  payable  quarterly, 
to  commence  in  9  yr.  10  mon.  18  da.,  int.  10  %,  payable 
semi-annually.  Ans.  $3858.88 

CASE   III. 

ART.  357.  Given,  the  rate,  the  payment,  the  interval, 
and  the  time  to  run,  to  find  the  present  value  of  an  an- 
nuity certain. 

RCLE.  —  Find  the  present  value  of  two  perpetuities  having  the 
given  rate,  payment,  and  interval,  one  of  them  commencing  when 
the  annuity  does,  and  the  other  when  the  annuity  ends.  The  dif- 
ference between  these  values  will  be  the  present  value  of  the 
annuity. 

NOTE.  —  This  rule  applies  whether  the  annuity  i«  immediate  or 
deferred;  in  the  latter,  the  time  the  annuity  is  deferred  must  be 
known,  and  used  in  getting  the  values  of  the  perpetuities. 

2.  By  using  the  initial  instead  of  the  present  values  of  the  perpe- 
tuities, the  rule  gives  the  initial  value  of  the  deferred  annuity,  which 
may  be  used  in  finding  its  final  or  forborne  value.  (Rein.  1,  Case  IV.) 

1.  Find  the  present  value  of  an  immediate  annuity  0' 
$250  continuing  8  years,  6  °f0  interest. 

Pres.  val.  of  immediate  perpetuity  of  J'tfO,  .  .  .  —  $4166.67 
Pres.  val.  of  perpetuity  of  $250,  deferred  8  yr.  .  .  =  2614.22 
Pres.  val.  of  immediate  annuity  of  $250,  ninn'nr  3  JT  --  $15r>2.l5 


REYIEW.—  356.    What  is  Case  2?     The  rule?     Solve  the   example. 
357.  What  is  Civso  3  ?     Thornlo? 


ANNUITIES.  30? 


2.  The  present  value  of  an  annuity  of  $080,  to  com- 
mence in  7  yr.  and  continue  10  yr.,  5  %  ir,t. 

1'res.  7al.  of  perpetuity  of  $080,  deferred  7  yr.,  at  5  %,      =  $9665.27 
Pres.  val.  of  perpetuity  of  $680,  deferred  17  yr.,  at  5  fc,    —    6933.20 

Pros.  val.  of  annuity  of  $680,  defer'd  7yr.,  to  run  10  yr.  =  $3732.07 

3.  Th6  pres.  val.  of  ah  annuity  of  $125,  to  commence 
in  12  yr.  and  run  12  yr.,  int.  7  %.  Arts.  $440.83 

4.  The  present  value  of  an  immediate  annuity  of  $400 
running  15  yr.  G  mon.,  int.  8  f-/0.  A»s.  $3484.41 

5.  The  pres.  val.  of  an  annuity  of  $826.50,  to  com- 
mence in  3  yr.  and  run  13  yr.  9  mon.,  int.  G  %,  payable 
scnii-annually.  Ans.  $0324. GO 

0.  The  present  value  of  an  annuity  of  $60,  deferred 
12  yr.  and  to  run  9  yr.,  int.  4i  %.  Ans.  $257.17 

7.  Sold  a  lease  of  8480  a  yr.,  payable  quarterly,  having 
8  yr.  9  mon.  to  run,  for  $2500 :  do  1  jrain  or  lose,  int.  8  %, 
payable  seuii-annually?  Ans.  Lose  $509.90 

CASE  IV. 

ART.  358.  Given,  the  payment,  the  interval,  the  rate, 
and  time  to  run,  to  find  the  final  or  forborne  value  of  an 
annually. 

RULE. — Consider  Hie  annuity  a  perpetuity,  and  find  its  initial 
value,  by  Cute  I.  The  compound  interest  <>f  tki*  yum,  at  the  git-en 
rate  for  the  time  the  annuity  ntus,  will  be  theflnal  or  forborne  value. 

Find  the  final  or  forborne  value  of  an  annuity  of  $250, 
continuing  8  yr.,  int.  G  %. 

SOLCTIOS. — The  initial  value  of  a  perpetuity  of  $2-jO,  at  6  %,  = 
$4166.00-|;  its  compound  interest,  at  0  %  for  8yr.,  =  $41CG/.r>2  x 
.6938481  =  $2474.37,  the  final  or  forborne  value  of  the  annuity. 

REMARKS. — 1.  The  final  or  forborne  value  of  an  annuity  may  be 
got  by  first  getting  its  initial  value  by  Case  III,  and  finding  its  com- 
pound amount  for  the  time  the  annuity  runs,  at  (he  given  rate. 

2.  The  present  value  of  an  annuity  can  be  obtained,  by  first  get- 
ting its  final  or  forborne  value  by  the  rule  in  this  case,  and  finding 
its  present  worth  for  the  time  the  annuity  runs,  at  the  given  rate. 


REVIEW. — 357.  What  kind  of  annuities  does  tlio  rule  apply  to?  IIcw 
can  the  initial  rnlue  of  a  deferred  annuity  be  found?  Of  what  uso  is  it? 
358.  What  is  Case  4?  What  is  the  rule  of  Case -1  ?  How  el.«e  may  the 
final  value  (if  nn  annuity  bo  found?  Iluw  may  the  present  valio  of  an 
annuity  bo  found  from  this  rule? 


308 


RAY'S    HIGHER  ARITHMETIC. 


ART.  859.  The  present  value  o/$l  per  annum  in  any  number  of  years. 


Yrs. 

4  per  cent. 

5  per  cent. 

6  per  c«nt. 

7  per  cent. 

8  percent. 

10  per  cent. 

1 

.901538 

.0523$! 

.943396 

.934579 

.92P.J20 

.909001 

2 

].8Sii095 

1.859410 

1.833393 

1.808018 

1.783265 

1.735537 

3 

2.775091 

2.723218 

2.073012 

2-024316 

2.577097 

2.480852 

4 

3.029895 

3.545951 

3.405100 

3.387211 

3.312127 

3.100805 

5 

4.451822 

4.329477 

4.212304 

4.100197 

3  992710 

3.790787 

6 

5.212137 

5.075092 

4.917324 

4.700540 

4.C22880 

4.355261 

7 

O.o;t;o.~>5 

5.780373 

5.582381 

5.389289 

6.2<i'i370 

4.808419 

8 

C.  732745 

0.403213 

C.  209794 

5.971299 

6.746639 

5-334026 

9 

7.435332 

7.107822 

C.  801092 

G.  515232 

G-  24(iS88 

5.759024 

10 

8.U08UG 

7  721735 

7.300087 

7.023582 

G.  710081 

G.  144507 

11 

8.700477 

8.300414 

7.880875 

7.498074 

7.138964 

6.495001 

12 

9.385074 

8.803202 

8.383844 

7.942080 

7.530078 

0.813092 

13 

9.085048 

9.393573 

8.852083 

8.357051 

7.903770 

7.103356 

14 

10  503123 

9.898041 

0.294984 

8.745408 

8.244237 

7.300087 

15 

11.118387 

10.379058 

9.712249 

9.107914 

8.559479 

7.00of80 

16 

11.052296 

10.837770 

10.105895 

9.446C49 

8-8513C9 

7.825709 

17 

12.1G5iJ(i9 

11.274000 

10.477200 

9.703223 

9.121G38 

8.021553 

IS 

12.G592!)7 

11-089587 

10.827003 

10.059087 

0.371887 

8-201412 

10 

13.  13:59.7.) 

12.085321 

11.108110 

10.335595 

9.003599 

8.304920 

20 

13.590320 

12.402210 

11.409921 

10.594014 

9.818147 

8.513504 

21 

14.029100 

12.821153 

11.704077 

10.835527 

10-010803 

8.G48G94 

22 

14.451115 

13.103003 

12.0415S2 

11.001241 

10.200744 

8.771540 

23 

14.856342 

13.488574 

12.30337!) 

11.272187  ' 

10.371059 

8.883218 

21 

15.2401)03 

13.798042 

12.550.*J8 

11-409334 

10.528758 

8.981744 

25 

15.022080 

14.093945 

12.783350 

11.053583 

10.074770 

9.077040 

20 

15.982709 

14.375185 

13-003100 

11.825779 

10.809978 

9.1G0915 

27 

10.32J58-3 

14.043034 

13.2105:;! 

11.980709 

10.9:50105 

s.atfma 

28 

10.0(33003 

14.898127 

13.400104 

12.137111 

11-051078 

9.300507 

29 

10.983715 

15.141074 

13.590721 

12.277074 

11.158400 

9.309000 

30 

17.292033 

15.372151 

13.704831 

12.40L/UU 

11.257783 

9.420914 

31 

17.588494 

15.592811 

13.929080 

12.531814 

11.349799 

9.479013 

32 

17.873552 

15.802077 

11-084043 

12.040505' 

11.434999 

9.520370 

33 

18.147040 

10.002.549 

14.230230 

12.753790 

11.513888 

9.609432 

34 

18.411198 

Hi.  192904 

14.308141 

12.854009 

11.580931 

9.608375 

35 

18.004013 

10.374191 

14.498210 

12-947072 

11.054508 

9.044159 

30 

18.908282 

10.  540852 

11.020987 

13.035208 

11.717193 

9.076508 

37 

19.ll2.-i79 

1G.7112S7 

14.730780 

13.117017 

11.77017:) 

9.705017 

38 

19.:!ii7804 

10.807893 

14.840019 

13.193173 

11.828809 

9.732051 

89 

19.581485 

17.017011 

14.949075 

13.204928 

11.878582 

0.750056 

10 

19.792774 

17.159080 

15.  0402117 

13-331709 

11.921013 

9.779051 

41 

19.993052 

17.2943GS 

15.138016 

13.394120 

11.907235 

9.799137 

42 

20.180027 

17.423208 

15.224543 

13.452440 

12.006099 

9.817397 

43 

20.370795 

17.545912 

15.300173 

13.  500902 

12.043240 

9.833998 

41 

20.5l.SS4l 

17.002773 

15.383182 

13.557908 

12.077074 

9.849089 

45 

20.720040 

17.774070 

15.455832 

13.000022 

12.108402 

9.802S08 

40 

20.884054 

17.880007 

15.524370 

13.050020 

12.137400 

9.875280 

47 

21-012930 

17.981010 

15.589028 

13.091008 

12.1G4237 

0.88!i018 

48 

21.  195131 

18.077158 

15.050027 

13.730474 

12.189130 

9-890920 

49 

21.311473 

18.108722 

16.707673 

13.760799 

12.212103 

9.000290 

60 

21.482185 

18.255925 

15.701801 

13.800740 

12.233485 

9.014811     ] 

51 

21.C174S5 

18.  338977 

15.813070 

13-832473 

12.203227 

9.922559 

52 

21.717582 

18.418073 

15.:  s  ;i:i'.u 

13.80-2121 

12.27.luUJ 

0.929399 

53 

21.  872075 

18.493403 

15-900971 

13.8S98.-iii 

12.288132 

9.9:55999 

51 

21.092957 

18.505146 

15.919970 

13.915735 

12.304103 

9.941817 

&5 

22.108012 

18.033472 

15.990543 

13.9399:!!) 

12.318014 

9.947107 

ANNUITIES.  309 


1.  Find  the  forborne  value  of  an  immediate  annuity  of 
5300,  running  18 yr.,  int.  5  %.  AHA.  $8439.72 

2.  A  pays  $25  a  year  for  tobacco;  how  much  better  off 
would  he  have  been  in  40  yr.,  if  he  had  invested  it  at  10% 
per  annum?  Ans.  $11004.81 

3.  Find  the  forborne  value  of  an  annuity  of  $75,  to 
commence  in  14  y*;.,  and  run  9yr.,  int.  6%.  Ans.  $801.85 

SUGGESTION. — The  14  yr.  is  not  used. 

4.  A  pays  $5  a  year  for  a  newspaper ;   if  invested  at 
9  %,  what  will  his  subscription  have  produced  in  50  yr.? 

Ans.  $4075.42 

5.  B   pays   $150  a   year,  to  have   his  life  insured  for 
$5000:  if  he  dies   in  20  yr.,  does  the  Insurance  Co.  gain 
or  lose?  (int.  6  %.)  Ans.  Gains  $517.84 

ART.  360.  By  the  table  just  given,  some  interesting  and 
important  cases  in  annuities  can  be  solved,  among  which 
are  the  following  three : 

CASE    V. 

Given,  the  rate,  time  to  run,  and  the  present  or  final 
value  of  an  annuity,  to  find  the  payment. 

RULE. — Assume  J$l  for  the  payment;  determine  the  present 
or  final  value  on  this  supposition,  and  divide  the  given  present 
or  final  value  by  it. 

NOTE. — Get  the  final  value,  if  necessary,  by  the  rule  in  Case  IV. 

1.  An  immediate  annuity  running  11  yr.,  can  be  pur- 
chased for  $6000:  what  is  the  payment,  int.  G  %  ? 

SOLUTION. — The  present  value  of  an  immediate  annuity  of  SI  for 
tly.-.,  at  0^,  is  $7.886875;  $6000  divided  by  this  gives  §700.76, 
the  payment  required. 

2.  How  much  a  year  should  I  pay,  to  secure  $15000 
it  the  end  of  17  yr.,  int.  7  %  ?  Am.  $486.38 

3.  What  is  the  payment  of  an  annuity,  deferred  4  yr., 
running  16  yr.,  and  worth  $4800,  int.  6%?  Ans.  $599.64 

CASE  VI. 

ART.  361.  Given,  the  payment,  the  rate,  .and  prescn 
value  of  an  annuity,  to  find  the  time  it  runs. 

lit* I.E. —  Divide  the  present  value  by  the  yearly  payment:  the 
quotient  will  be  the  present  value  of  an  annuity  of  81,  similar  to 
\he  one.  proposed ;  by  looking  in  the  table  for  this  number,  under 


RAY'S   HIGIIElt   ARITHMETIC. 


the  head  of  the  given  rate  per  cent.,  ike  corresponding  number  of 
years  can  be  J'uund. 

NOTE. — If  the  number  is  not  in  the  table,  take  the  nearest  one 
below  it ;  the  balance  remaining  can  be  ascertained  by  finding  the 
difference  between  the  compound  amount  of  the  present  value,  and 
the  final  amount  of  the  annuity  for  the  number  of  yeais  taken  out. 

REMARK. — This  case  is  of  use  in  finding »the  time  required  to 
C  puidate  a  debt  drawing  interest,  by  means  of  a  sinking  fund,  that 
is,  by  installments  at  regular  intervals  of  a  year,  half-year,  or 
quarter  of  a  year;  for  example, 

In  what  time  will  a  debt  of  $10000,  drawing  interest 
at  6  5^,  be  paid  by  installments  of  §1000  a  year? 

SOLUTION. — The  $10000  may  be  considered  the  present  value  of 

an  annuity  of  $1000  a  year  at  6  %;  but  $10000  -f-  $1000,  =  $10,  the 

present  value  of  an  annuity  of  $1  for  the  same  time  and  rate ;  by 

reference  to  the  table,  the  time  corresponding  to  this  present  value, 

under  the  head  of  6  %,  is  15 yr.,  a  balance  being  then  due  of  $1)89.01. 

Comp.  amt. of  $10000  for  15 yr., at 6  ^,  (Art.  348.)     .     =$23905.58 

Final  val. of  annuity  $1000  for  15yr.,  at  6^,(Art.358.)  =    23275.97 

Bal.  due  at  end  of  15  yr $689.61 

1.  In  how  many  years  can  a  debt  of  $1000000,  drawing 
interest  at  6  %,  be  discharged  by  a  sinking  fund  of  $80000 
a  year?  .4ns.  23  yr.,  and  $60083.43  then  unpaid. 

2.  In    how  many   years    can    a   debt   of  $30000000, 
drawing  interest  at   5  %,  be    paid   by  a    sinking   fund  of 
$'2000000?  Am.  28  yr.,  and  $798709.00  unpaid. 

3.  In  how  many  years  can  a  debt  of  $22000,  drawing 
7  %  interest,  be  discharged  by  a  sinking  fund  of  $2500  a 
year?  Ans.  14  yr.,  and  $351.53  then  unpaid. 

CASE   VII. 

ART.  362.  Given,  the  payment,  time  to  run,  and  present 
value  of  an  annuity,  to  find  the  rate  of  interest. 

RCLE. — Divide  the  present  value  by  the  yearly  payment ;  the 
quotient  will  be  the  present  value  of  an  annuity  of  $1  similar 
to  the  one  proposed;  by  looking  in  the  table  opposite  the  given 

R  K  v  i  K  w.— SCO.  What  is  Case  5  ?  The  rule  ?  How  may  the  final  valua 
Df  the  annuity  of  $1  be  found  if  necessary?  8til.  What  is  Cnso  *5  ?  What 
Is  the  rule  of  Ca.se  C  ?  If  the  number  of  intervals  bo  not  exact,  how  find 
the  balance  remaining  for  the  partial  interval  ? 


CONTINGENT   ANNUITIES.  31  I 

number  of  years  until  this  number  is  found,  the  rale  per  cent,  at 
the  head  <>f  Ike  column  which  contains  it,  will  be  the  rate  required. 

1.  If  $9000  is  paid  for  an  immediate  annuity  of  $750 
to  run  20  yr.,  what  is  the  rate?  Ans.  About  5^  %. 

2.  If  an    immediate   annuity  of  $80,   running   14  yt, 
sells  for  $(J50,  what  is  the  rate?  Ans.  8 


CONTINGENT   ANNUITIES. 

ART.  363.  Contingent  Annuities  comprise  Life  Insur- 
ance, Dowers,  Pensions,  &c. 

A  person  who  has  his  life  insured  pays  a  fixed  sura, 
called  the  annual  premium,  every  year  he  lives,  and  at  his 
death,  the  sum  insured  is  paid  to  the  person  designated 
in  the  policy. 

The  premium  is  a  certain  rate  per  cent,  of  the  sum  in- 
sured, varying  with  the  age  of  the  applicant. 

Rills  of  Mortality  are  tables  showing  how  many  of  a  given 
number,  say  10000,  born  in  the  same  year,  die  at  each  age. 

Expectation  of  life  is  the  average  number  of  years  a 
person  of  any  age  lives,  as  deduced  from  bills  of  mortality. 

A  table  of  life-annuities  shows  the  sum  to  be  paid  by  a 
person  of  any  age,  to  secure  an  annuity  of  $1  for  life. 

CASE  I. 

ART.  384.  To  find  the  value  of  a  given  annuity  on  the 
life  of  a  person  whose  age  is  known. 

RPI,E. — Find  from  the  table  the  value  of  a  life-annuity  of  §1, 
for  the  giren  age  and  rate  of  interest,  and  multiply  it  by  the  pay- 
ment of  the  given  annuity. 

NOTES. — 1.  To  find  the  value  of  a  life-estate  or  widow's  dowei 
(which  is  a  life-estate  in  one-third  of  her  husband's  real  estate)' 
Estimate  the  value  of  the  property  in  which  the  life-estate  is  held;  tht 
yearly  interest  of  this  sum  at  an  agreed  rate,  will  be  a  life-annuity,  whose 
value  for  the  given  age  and  rate,  will  be  the.  value  of  the  life-estafe. 

2.  The  reversion  of  a  life-annuity,  life-estate,  or  dower,  is  found  by 
deducting  its  value  from  the  value  of  the  property. 

R  r.  v  i  B  w.— 362.  What  is  Case  7?  The  rule?  363.  What  do  contin- 
gent annuities  comprise?  What  is  life  insurance?  The  annual  premium! 
The  policy?  How  is  the  premium  estimated?  What  are  bills  of  uior 
tality?  What  5?  espectntion  of  life?  What  is  a  table  of  life-annuities? 


312 


RAY'S   HIGHER  ARITHMETIC. 


TABLE 

Showing  the  values  of  Annuities  on  Stnyle  Lives,  according  to  iJie 
Carlisle  Table  of  Mortality. 


jAije. 

I  por  cent. 

5  per  Ct. 

j  per  ct. 

7  j'cr  ct. 

A;e. 

1  per  cent. 

j  per  ct. 

0  per  ct. 

7  per  ct. 

C 

14.28104 

12.083 

10.439 

9.177 

62 

12.25703 

11.154 

10.208 

9.302 

1 

10.65155 

13.995 

12.078 

10.MI5 

63 

11  01503 

10.8W 

9.988 

9.2(1.-) 

8 

17.72ulO 

14.983 

12.025 

11.312 

64 

11.02C73 

10.021 

0.7.11 

9.011 

3 

IS.  7  1508 

15.824 

13.052 

11.078 

65 

11.2001:1 

i^.:;i7 

9.52) 

8.807 

4 

1J.  23133 

10.271 

14.042 

12.322 

60 

10.00007 

10.003 

0.2SO 

8.69J 

5 

19.50203 

1G-500 

14.325 

12.574 

67 

10.C2550 

9.771 

9.027 

8.375 

G 

VJ  74602 

10.735 

14.4uO 

12.f/.»8 

68 

10.28047 

9.478 

8.772 

8.15.1 

7 

J'J.70019 

10.7:10 

14.518 

12.750 

63 

9.9!i33l 

0.10'.) 

8.52J 

7.  WO 

8 

)0.7^U.J 

1G.78J 

n.52j 

12.77  J 

CO 

9.oo:Ky 

8.910 

8.301 

7.74:j 

9 

l'J.OUU4 

10.742 

14.500 

12.7->4 

01 

9.  3980  J 

8.712 

8.108 

7.572 

10 

1'J.  58339 

10.GCO 

14.448 

12.717 

C2 

9.13G7G 

8.487 

7.913 

7.403 

11 

10.45o57 

10.581 

14.3al 

12.0u9 

63 

8.8715.) 

8.258 

7.711 

7.22.) 

12 

19.3-J4J3 

10.494 

14.321 

12.021 

04 

8.5:w:;o 

8.010 

7.602 

7.0J2 

13 

10.20037 

lii.4o:5 

14.207 

12.572 

Go 

8.30719 

7.7  19 

7.281 

0.8^17 

U 

1J.  06182 

10.310 

14.191 

12.5ii 

00 

8.00000 

7.503 

7.040 

0-0-11 

\ 

IS 

18.05531 

10.227 

11.125 

12.473 

67 

7.09980 

7.227 

G.803 

G.421 

16 

18.83030 

10.114 

14.0o7 

12.420 

08 

7.37070 

0.041 

0.540 

O.UJ 

17 

IS.  72111 

10.000 

14.012 

12.38!) 

09 

7.04881 

0.043 

0.277 

5  945 

18 

18.00060 

15.987 

13.950 

12.318 

70 

0.70030 

8.336 

5.998 

5  COO 

1;) 

18.48,^0 

15.004 

13.8J7 

12.305 

71 

0.35773 

0.0i5 

5.704 

5.420 

20 

18.30170 

15.817 

13.835 

12.259 

72 

C.  02518 

5.711 

5.424 

5.1G2 

21 

18.2..196 

15.72o 

13.7(3 

12.210 

73 

5.72405 

5.435 

5.170 

4.H27 

22 

18.00380 

15.028 

13.007 

12.150 

74 

5.45812 

5.100 

4.944 

4.719 

83 

17.05010 

15.525 

13.021 

12.008 

75 

6.23001 

4.980 

4.700 

4.510 

2i 

17.80058 

15.417 

13.5U 

12.037 

70 

5.02399 

4.702 

4.579 

4  382 

2.-> 

17.G448C 

15.303 

13.450 

11.972 

77 

4.82473 

4.  009 

4  410 

4  227 

2,i 

17.46580 

15.187 

13.308 

11.5101 

78 

4.02100 

4.422 

4.238 

4  0;i7 

27 

17.32023 

15.005 

13.275 

11.832 

70 

4.30345 

4.210 

4.040 

3.883 

23 

17.15412 

14.942 

13.182 

H.75J 

81) 

4.  -18280 

4.015 

3.8-58 

3.713 

2!) 

10.00083 

14.827 

13.00J 

11.093 

81 

3.95300 

3.790 

3.050 

3.523 

30 

10.85215 

14.723 

13.020 

11.C3G 

82 

3.74534 

3.  COG 

3.474 

3.352 

81 

10.70511 

14.017 

12.942 

11.578 

KJ 

3.534O9 

3.400 

3.280 

3.174 

82 

10.65240 

14.603 

12.800 

11.610 

84 

3.32850 

3.2il 

3.102 

2.000 

33 

10.39072 

14.387 

12.771 

11.448 

85 

3.1151.-, 

3.00'J 

2.900 

2.815 

31 

10-21043 

14.200 

12.075 

11.374 

80 

•2.92831 

2.S30 

2.739 

2.052 

35 

1G.04123 

14.127 

12.573 

11-295 

J'7 

2.77593 

2.C85 

2.590 

2.510 

36 

15.85.J77 

13.987 

12.405 

11.211 

88 

2.08337 

2.597 

2.515 

2.430 

37 

15.00580 

13.843 

12.354 

11.124 

8J 

2.67704 

2.495 

2.417 

2.344 

38 

15.47120 

13.095 

12.23'J 

11.033 

Ul) 

2.41021 

2.339 

2.20(5 

2.108 

39 

15  27181 

13.542 

12.120 

10.9JO 

SA 

2.39335 

2.321 

2.248 

2.180 

40 

1.1.07303 

13.390 

12.002 

10.845 

92 

2.49199 

2.412 

2.337 

2.200 

•11 

14.88314 

13.245 

11-800 

10.757 

»J 

2.59955 

2.518 

2.4K) 

2.:  :i,.7 

42 

n.<;(»4»;o 

13.101 

11-779 

10.071 

1>4 

2.C4976 

2.500 

2-402 

2.410 

43 

14.50529 

12.957 

11.  CCS 

10.585 

95 

2.67433 

2.698 

2.622 

2.451 

44 

14.30874 

12.800 

11.551 

10.494 

90 

2.02770 

2.555 

2.480 

2.420 

45 

14.10400 

12.048 

11.428 

10.397 

07 

2.49201 

2.428 

2.3.18 

2.300 

40 

13.88928 

12.480 

11.290 

10.202 

98 

8.2T8 

2   177 

47 

13.00208 

12.301 

11.154 

10.178 

B9 

2.0S700 

2.015 

2  00V 

1.994 

48 

13.419U 

12.107 

10.998 

10.052 

100 

1.C5282 

1.024 

i-eoo 

1.5ii9 

49 

13.15312 

11.892 

10.823 

9.908 

101 

i.2ioa-. 

1.192 

1.175 

1.150 

60 

12.86002 

11.000 

10.631 

9.749 

102 

0.7G183 

0.753 

0.744 

0.735 

61 

12.60681 

11.410 

10.422 

9.573 

in 

0.32051 

0.317 

0.314 

0.312 

CONTINGENT  ANNUITIES.  318 


1.  What  must  be  paid  for  a  life-annuity  of  $650  a  year, 
by  a  person  aged  72  yr.,  int.  7  %1  Ans.  $3355.30 

2.  What  is  the  life-estate  and  reversion  in  $25000,  age 
55 yr.,  int.  6  %?  Ans.  Life-estate,  $14286;  Rev.  $10714. 

3.  The  dower  and  reversion  in  $46250,  age  21  yr.,  int. 
G#?  Ant.  Dower,  $12736.33;  Rev.  $2680. 34 

CASE   II. 

ART.  365.  To  find  how  large  a  life-annuity  can  be  pur- 
chased for  a  given  sum,  by  a  person  whose  age  is  known. 

RULE.  —  Assume  $1  a  year  for  the  annuity;  fold  from  the 
table  its  value  for  the  given  age  and  rate  of  interest,  and  divide  the 
given  cost  by  it :  the  quotient  will  be  the  payment  required. 

How  large  an  annuity  can  be  purchased — 

1.  For  $500,  age  26  yr.,  int.  6  %  ?        Ans.  $37.40 

2.  For  $1200,  age  43,  int.  5  %  ?  Ans.  $92.61 

3.  For  $840,  age  58,  int.  7  %  ?  Ans.  $103.03 

CASE   III. 

ART.  366.  To  find  the  present  value  of  the  reversion  of  a 
given  annuity ;  that  is,  what  remains  of  it,  after  the  death 
of  its  possessor,  whose  age  is  known. v 

RULE. — Find  the  present  value  of  the  annuity  during  its  whole 
continuance ;  find  its  value  during  flie  given  life :  their  difference 
Kill  be  the  value  of  the  reversion. 

NOTE. — It  will  save  work,  to  consider  the  annuity  as  $1  a  year,  then 
apply  the  rule,  using  the  tables  in  Art.  359  and  Art.  364,  and  multiply 
the  result  by  the  given  payment. 

1.  Find  the  present  value  of  the  reversion  of  a  perpe- 
tuity of  $500  a  year,  after  the  death  of  a  person  aged  47, 
int.  5$.  Ans.  $3849.50 

2.  Of  the  reversion  of  an  annuity  of  $165  a  year  for 
30  yr.,  after  the  death  of  a  person  38  yr.  old,  int.  6  %. 

Ans.  $251.76 

3.  Of  the  reversion  of  a  lease  of  $1600  a  year,  for  40  yr 
after  the  death  of  A,  aged  62,  int.  7  %.  Ans.  $9485.93 

CASE  IV. 

ART.  367.  To  find  the  single,  and  annual  premium,  paid 
by  one  of  a  given  age,  to  secure  a  given  sum  at  death. 

REVIEW.— 364.  What  is  Case  1?     The  rule?     365.  What  is  Case  27 
Tho  rule?    366.  What  is  Case  3?    The  rule: 
27 


814  RAY'S   HIGHER   ARITHMETIC. 


RULE.— .F^'d  the  present  value  of  an  immediate  perpetuity  of 
51,  at  the  yiven  rate ;  subtract  from  it  the  present  value  of  an  an- 
nuity of  Si  for  the  given  age  and  rate ;  divide  the  remainder  by 
the  calue  of  the  perpetuity  increased  by  1:  the  quotient  will  be  the 
present  value  of  §1  due  at  the  expiration  of  the  given  life,  and  if 
it  be  multiplied  by  the  given  sum,  the  product  will  be  the  single 
premium. 

To  get  the  annual  premium,  divide  the  single  premium  by  1  more 
han  the  present  value  of  an  annuity  of  $1  for  Vie  given  age  and 
rate. 

NOTE. — The  annual  premiums  being  paid  in  advance,  the  single 
premium  must  exceed  the  present  value  of  an  ordinary  life-annuity 
of  the  same  payments,  by  one  payment,  viz :  the  one  made  when  the 
insurance  is  obtained. 

1.  Find  the  single  premium  and  annual  premium  ne- 
cessary to  secure  §1000  at  the  death  of  a  person  aged  32, 
allowing  5  %  interest.  Ans.  $261.62,  and  $16.87 

2.  To  secure  $1500  at  the  death  of  a  person  aged  66, 
allowing  6  %  interest.         Ans.  $816.59,  and  $101.45 

CASE  V. 

ART.  368.  If  the  holder  wishes  to  sell  his  policy,  or  sur- 
render it  to  the  Insurance  Co.,  proceed  thus, 

TO  VALUE  A  POLICY  OF  LIFE  INSTTRANCE, 

RULE. — Take  the  difference  beticeen  the  premium  in  the  policy, 
and  the  premium  for  insuring  the  holder's  life  for  the  same  sum, 
now ;  multiply  it  by  1  more  than  the  present  value  of  a  life-annuity 
of  §1  on  the  present  age;  this  product  will  be  the  value  of  the 
policy. 

1.  The  premium  in  the  policy  is  $2.04  per  $100  ;  the 
premium  now,  $2.50  per  $100;  age  27 yr.,  int.  6^:  value 
the  policy  for  $5000.  Ans.  $328.33 

2.  Value  the  policy,  when  the  premium  in  it  is  $74,  thii 
premium  now,  $82,  age  42  yr.,  int.  5  %.  Ans.  $112.81 

3.  When  the  premium  in  the  policy  is  $51.76,  the  pre- 
mium now,  $68. 44,  age  36  yr.,  int.  7  %.  Ans.  $203.68 

R  E  TI  E  w. —  367.  What  is  Case  4  ?  The  rule  ?  How  does  life-insuranc« 
differ  from  ordinary  life-annuities  ?  Hfl8.  Give  the  rule  for  valuing  a  polk? 
of  li£p-inpurano.o. 


PROPORTIONAL  PARTS.  315 


XXXI.    PROPORTIONAL  PARTS. 

ART.  369.  Divide  the  number  90  into  3  other  numbers 
proportional  to  2,  3  and  4. 

SOLUTION. — Since  the  required  numbers  are  to  have  the  same 
ratios  as  2,  3  and  4,  they  must  contain  a  common  factor  2,  3  and  4 
times  respectively.  Call  this  common  factor  a  part;  then  the  1st 
number  is  2  parts,  the  2d  is  3  parts,  the  3d  is  4  parts;  and  the  whole 
number  90  =  2  +  3  +  4,  or  9  parts.  Hence,  1  part  is  ^  of  90  =  10, 
and  the  1st  number  is  §  of  90  =  20,  the  2d  is  |  of  90  =  30,  and  the 
3d  is  |  of  90  =  40. 

PROOF.— 20  +  30  +  40  =  90;  and  20,  30  and  40  have  the  same 
ratios  as  2,  3  and  4,  omitting  the  common  factor  10. 

TO  DIVIDE  A  NUMBER  INTO  PROPORTIONAL  PARTS, 

RULE. —  Take  separately  sucJi  parts  of  the  number  to  be  divided, 
as  each  of  the  proportional  numbers  is  of  their  sum ;  the  results 
will  be  the  required  numbers. 

PROOF. — The  sum  of  the  parts  must  be  equal  to  the  number 
divided,  and  the  ratio  of  each  to  its  proper  proportional  must  be 
the  same. 

NOTES. — 1.  If  the  proportional  numbers  are  fractional,  reduce 
them  to  a  least  common  denominator,  and  use  their  numerators  for 
the  proportional  numbers. 

2.  Any  common,  factor  may  be  canceled  out  of  the  proportional 
numbers,  before  using  them. 

1.  Divide  the  number  84  into  2  parts  proportional  to 
30  and  40.  -4ns.  36,  48. 

Suo. — For  30  and  40,  use  3  and  4,  dropping  the  common  factor  10. 

2.  Divide  60  apples  among  3  boys  in  proportion  to  their 
ages,  which  are  T,  10  and  13.  An*.  14,  20,  26. 

3.  One  part  of  a  pole  16  ft.  long,  is  f  of  the  other  part : 
find  each  part.  Ans.  6  ft.,  and  10  ft. 

Suo. —  f  and  |,  or  3  and  5  are  the  proportional  numbers. 

4.  i  of  A's  money  is  equal  to  f  of  B's,  and  both  have 
$222  j  how  much  has  each?  Ans.  A  $162,  B  $60. 

SUGGESTION. — A's=  f  J  °f  B's;  if  B's  is  10  parts,  A's  is  27  parts: 
hence,  27  and  10  are  the  proportional  numbers. 

REVIEW. — 369.  What  is  tho  rule  for  dividing  a  number  into  proper- 
tional  parts?  Tho  proof?  If  the  proportional  numbers  aro  fractional, 
what  should  be  done?  If  they  are  large?  Solve  tho  example. 


316  RAYS    HIGHER   ARITHMETIC. 


5.  A  tree,  51ft.  high,  was  broken  by  the  wind;  §  of  the 
part  that  fell  was  equal  to  4  of  the  stump:  how  long  was 
each  ?  Ans.  Stump  24  ft.,  other  part  27  ft. 

6.  Divide  38  in  proportion  to  4g  and  3|  Ans.  20, 18. 

Sou. — 4g  and  3-|  are  equal  to  2g5  and  l^s,  or  f£  anc*  1§>  hence,  50 
and  45 — or,  dividing  by  5 — 10  and  9  are  the  proportional  numbers. 

7.  Divide  the  cost  of  a  supper,  $4.50,  among  3  persons 
to  proportion  to  their  money:  A  has  $100,  B  $75,  C  $125. 

Ans.  A  $1.50,  B  $1.12A,  C  $1.87^ 
8     A  father  left  $7140  to  his  children,  A,  B  and  C;  A's 
share  was  to  B's  as  2  to  3,  and  B's  to  C's  as  4  to  5 :  what 
did  each  get?          Ans.  A  $1632,  B  $2448,  C  $3060. 

SUGGESTION. — If  C's  share  is  5  parts,  B's  is  4  parts,  and  A's  is 
|  of  B's,  that  is,  f  of  4  parts  =  2|  parts ;  hence,  2f ,  4  and  5,  or  f , 
S2'  Vi  or  8>  12,  and  15,  are  the  proportional  numbers. 

9.  A  paid  $3.60,  and  B,  $2  for  a  bbl.  of  flour  (196  lb.): 
how  many  lb.  should  each  have  ?   Ans.  A  126  lb.,  B  70  lb. 

10.  Distribute  $180  among  5  poor  families  in  proportion 
to  the  number  of  children  in  each,  which  are  3,  4.  5,  2,  6 

Ans.  1st  $27,  2d  $36,  3d  $45,  4th  $18,  5th  $54. 

11.  Divide  1065  in  proportion  to  3,  5  and  7;  also  in  pro- 
portion to  i  i,  i  Ans.  213, 355, 497;  and  525,  315,  225. 

12.  How  much  copper  and  tin,  100  parts  of  the  former 
to  11  of  the  latter,  will  make  a  cannon  weighing  18cwt. 
3  qr.  12  lb.?     Ans.  17cwt.  copper,  1  cwt.  3  qr.  12  lb.  tin, 

13.  U.  S.  standard  silver  and  gold  are  9  parts  pure  metal 
to  1  part  alloy:  how  much  pure  silver  in  the  half-dollar,  which 
weighs  8  pwt.?  how  much  pure   gold  in   the  eagle,  which 
weighs  10  pwt.  18gr.? 

Ans.  7 pwt.  4.8gr.,  and  9  pwt.  16.2gr. 

14.  Pewter  is  112  parts  tin,  15  lead,  and  6  brass:  how 
much  of  each  ingredient  in  2  lb.  1  oz.  4  dr.  of  pewter? 

Ans.  lib.  12 oz.  tin;   3oz.  12 dr.  lead;  loz.  8  dr.  brass. 

15.  A  man  bequeaths  $875  to  A;  $2450  to  B;  $6035 
o  C:  if  the  estate  yields  only  $8190,  what  will  each  get? 

Ans.  A  $765. 62i,  B  $2143.75,  C  $5280.62^ 

16.  I  owe  $840,  due  Oct.  1 ;  I  pay  part  Aug.  15  (47  da. 
before-due),  the  rest  Jan.  1  (92  da.  after  due):  what  are  the 
payments?     Ans.  $555.97,  Aug.  15;  $284.03,  Jan.  1. 

BEUABK. — Divide  $840  in  proportion  to  47  and  92. 


PARTNERSHIP.  3  J  7 


17.  C  owes  $1200,  due  Nov.  6;    he  pays  part  Aug.  1, 
%nd  the  rest  Jan.  15;  what  are  the  payments? 

Am.  $502.99,  Aug.  1;  $69*7.01,  Jan.  15. 

18.  A  owes  $1680,  due  July  18;  he  pays  $480  before, 
the  rest  after  due;  when  were  the  payments  made,  if  they 
were  49  days  apart?  -4ns.  June  13,  and  Aug.  1. 


PARTNERSHIP. 

ART.  370.  An  important  application  of  Proportional 
Parts,  is  in  dividing  the  gains  or  losses  of  partners. 

Partners  are  persons  who  join  to  carry  on  a  business, 
and  constitute  a  firm,  house,  or  company. 

The  money  used  in  the  business,  is  called  the  capital  or 
stock,  and  is  contributed  by  the  partners. 

The  gain  or  loss  in  the  business,  is  called  the  dividend, 
because  it  is  to  be  divided. 

REMARK. — When  each  partner's  stock  is  employed  the  same 
time,  it  is  sometimes  called  Simple  Fellowship,  and  when  they  are 
employed  for  different  times,  it  is  called  Compound  Fellowship. 

CASE  I. 

To  divide  the  gain  or  loss,  when  each  partner's  stock  is 
employed  for  the  same  time. 

RULE. — Divide  the  gain  or  loss  among  the  partners  in  pro- 
portion to  their  shares  of  the  stock. 

A,  B  and  C  are  partners,  with  $3000,  $4000,  and  $5000 
stock,  respectively;  if  they  gain  $5400,  what  is  each  one's 
share? 

SoL.-Each       3     T2  of  $5400  =  $1350  A's  share. 
Bhouldhavethe        4      &  of  $  5400  =  $1  80  0  B's       „ 
name    part    of     _5      1*3  of  $  5400  =j^2J2j>0  C's        „ 

the  gain  as  ho     J^  ~$5~4~0  6  whole  gain, 

has  of  the  stock; 

hence,  the  gain  shoulJ  be  divided  in  proportion  to  their  stocks,  using 
8,  4,  5,  for  3000,  4000,  6000. 

REVIKW.— 370.  What  are  partners  7  What  is  the  firm?  What  is  the 
itock  or  capital  ?  The  dividend  ?  Why  so  called  ?  What  is  simple  fel- 
lowship ?  Compound  fellowship  ?  What  is  Case  1  of  partnership  ?  The 
rule?  Solve  the  example. 


318  RAY'S   HIGHER   ARITHMETIC. 

1.  A  and  B  gain  in  1  year  $3600;  their  store  expenses 
are  $1500.     If  A's  stock  was  $2500  and  B's  $1875,  how 
much  does  each  gain?  Ans.  A  $1200,  B  $900. 

Take  out  the  expenses,  then  divide. 

2.  A,  B  and  C  are  partners;  A  puts  in  $5000,  B  $6400 
C  $1600.     C  is  allowed  $1000  a  year  for  personal  attention 
to  the  business;  their  store  expenses  for  1  year  are  $800, 
and  their  gain,  $7000.     Find  A's  and  B's  gain,  and  C  « 
income.  Ans.  A  $2000,  B  $2560,  C  $1640. 

3.  A  pasture  rents  for  $160;  A  puts  in  24  cattle;  B, 
20;  C,  60;  D,  96;  and  they  pay  in  proportion:  what  doe  a 
each  pay?      Ans.  A  $19.20,  B  $16,  C  $48,  D  $76.80 

4.  A  and  B  speculate  together  in  flour;  A  contributes 
800  bbl.  at  $5.40  per  bbl.,  and  B,  600  bbl.  at  $4.80  per 
bbl. ;  they  lose  75  ct.  a  bbl.,  and  pay  storage  $68.45  :  what 
is  the  loss  of  each?  Ans.  A  $671.07,  B  $447.38 

5.  A,  B  and  C  form  a  partnership;  A  puts  in  $24000, 
B  $28000,  C  $32000;  they  lose  k  of  their  stock  by  a  fire, 
but  sell  the  remainder  at  f  more  than  cost:  if  all  expenses 
are  $8000,  what  is  the  gain  of  each? 

Ans.  A  $5714.28$,  B  $6666.663,  C  $7619.'04if 

6.  A,  B  and  C  are  partners:  A's  stock  is  $5760,  B's, 
$7200;  their  gain  is  $3920,  of  which  C  has  $1120:  what 
is  C's  stock,  and  A's  and  B's  gain? 

Ans.  C's  stock,  $5184;  A's  gain,  $1244. 44?;  B's  gain, 
$1555. 55§ 

7.  A,  B  and  C  aro  partners:    A's  stock,  $8000;   B's, 
$12800;  C's,  $15200;  A  and  B  together  gain  $1638  more 
than  C:  what  is  the  gain  of  each? 

Ans.  A  $2340;  B  $3744;  C  $4446. 

8.  A  and  B  buy  a  house  and  lease;  A  contributes  $3480; 
B,  $2900:  the  ground-rent,  taxes,  <tc.,  arc  $108.30,  and 
he  property  rents  for  $915.80:  what  docs  each  get? 

Ans.  A,  $440. 45 fT;  B,  $367.04TfiT 

9.  A,  B  and  C,  in  partnership,  have  capitals  respectively 
$19200,  $24000,  and  $32400;  they  sell  out  for  $100000; 
how  much  of  this  does  each  get? 

Ans.  A$25396.82|j;  B $31746. 03^;  C$42857.14? 

10.  Four  men  rent  57  A.  2  K.  16  P.  of  land  at  $3.75  an 
acre:  A  puts  in  72  sheep:  B,  80;  C,  96;  D,  112:  what  should 
each  pay?  Ans.  A$43. 20;  B$48;  C$57.60;  D$67. 20 


PARTNERSHIP   WITH  TIME.  310 

11.  A'B  gain  is  $1*750;  B's,  $1225;  C's,  $2275:  what 
part  of  the  capital,  $18690,  has  each? 

Ann.  A  $6230;  B  $4361;  C  $8099. 

12.  A,  B  and  C  have  a  joint  capital  of  $27000  :  neither 
draws  from  the  firm,  and  when  they  quit,  A  had  $20000; 
B,  $16000;  C,  $12000:  what  did  each  contribute? 

Ans.  A  $11250;  B  $9000;  C  $6750. 

CASE  II. — PARTNERSHIP  WITH  TIME. 

ART.  371  The  rule  for  distributing  gain  or  loss  among 
partners,  (Art.  370,)  applies  only  when  each  partner's 
capital  is  employed  for  the  same  time;  it  needs  some  modi- 
fication if  such  be  not  the  case. 

A,  B  and  C  are  partners:  A  puts  in  $2500  for  8mon., 
B,  $4000  for  6mon.;  C,  $3200  for  lOrnon.;  their  net 
gain  is  $4750:  divide  the  gain. 

SOLUTION. — A's  capital  ($2500),  used  8  months,  is  equivalent  to 
8  X  $2500,  or  $20000,  used  1  month ;  E's  capital  ($4000),  used  6 
months,  is  equivalent  to  6  X  $4000,  or  $24000,  used  1  month ;  C's 
capital  ($3200),  used  10  months,  is  equivalent  to  10  X  $3200,  or 
$32000  used  1  month.  Dividing  the  gain  ($4750)  in  proportion  to 
the  stock  equivalents,  $20000,  $24000,  $32000,  used  for  the  same 
time  (1  mouth),  the  results  will  be  the  gain  of  each. 

Ana.  A's  $1250,  B's  $1500,  C's  $2000. 

The  stock-equivalents  are  obtained  by  multiplying  each 
partner's  stock  by  the  time  it  is  used ;  hence, 

WHEN  THE  STOCKS  ARE  USED  FOR  DIFFERENT  TIMES, 

RULE. — Multiply  each  partner's  stock  by  the  time  it  is  used;  and 
divide  the  gain  or  loss  in  proportion  to  the  products  so  obtained. 

PROOF. — The  sum  of  the  separate  gains  or  losses  must  equal 
the  whole  gain  or  loss,  and  tho  ratio  of  each  partner's  gain  or 
loss  to  his  stock-equivalent  must  be  tho  same. 

NOTE. — Express  the  times  in  the  same  denomination,  before  mul- 
tiplying. 

1.  A  begins  business  with  $6000:  at  the  end  of  6  mon. 
he  takes  in  B,  with  $10000:  6  mon.  after,  their  gain  is 
$3300:  what  is  each  share  ?  Ana.  A's  $1800;  B's  $1500. 

R  E  v  r  K  w.— 371.  What  is  Case  2  7  What  are  stock  erinimleuts  ?  Thi 
rule?  The  proof?  In  multiplying  by  the  times,  what  ia  necessary? 


820  RAY'S   HIGHER    ARITHMETIC. 

2.  A  and  B  arc  partners:  A  puts  in  $2500;  B,  §1500. 
after  9  inon.,  they  take  in  C  with  $5000;  9  mon.  after,  their 
gain  is  $3250:  what  is  each  one's  gain? 

Ans.  A's  $1250;  B's  $750;  C's  $1250. 

3.  A  and  B  rent  a  pasture  for  $275 :  A  puts  in  80  sheep, 
and  B,  100;   after  6  mon.  they  each  sell  half  their  stock, 
and  allow  0  to  feed  50  sheep  the  rest  of  the  year:  how 
much  should  each  pay? 

Ans.  A  $103. 12i;  B  $128. 90|;  C  $42.96! 

4.  A  and  B  are  partners,  each  contributing  $1000  :  after 

3  months,  A  withdraws  $400,  which  B  advances;  the  same 
is  done  after  3  months  more;  their  year's  gain  is  $800: 
what  should  each  get?  Ans.  A  $200,  B  $600. 

5.  A  Avorks  9  hours  a  day;  B  remains  idlo  the  first  two 
days  of  the  week,  and  works  5|,  85,  10f  and  11 2  hr.  respec- 
tively, on  the  other  four:  if  their  week's  wages  are  $38.75, 
what  should  each  have?         Ans.  A  $23.25,  B  $15.50 

6.  A,  B  and  C  are  employed  to  empty  a  cistern  by  two 
pumps  of  the  same  bore:  A  and  B  go  to  work  first,  making 
37  and  40  strokes  respectively  a  minute;  after  5  minutes, 
each  makes  5  strokes  less  a  minute;  after  10  minutes,  A 
gives  way  to  C,  who  makes  30  strokes  a  minute  until  the 
cistern  is  emptied,  which  was  in   22  min.  from  the   start: 
divide  their  pay,  $2.         Ans.  A  46  ct, ;  B  $1 . 06 ;  C  48  ct. 

7.  A  and  B  are  partners:  A's  stock  is  to  B's,  as  4  to  5 : 
after  3  mon.,  A  withdraws  f  of  his,  and  B  f  of  his:  divide 
their  year's  gain,  $1675.  Ans.  A  $800,  B  $875. 

8.  A,  B  and  C  join  capitals,  which  are  as  -$,  3,  } :  after 

4  mon.,  A  takes  out  3  of  his;  after  9  mon.  more,  their  gain 
is  $1988:  divide  it.        Ans.  A  $714,  B  $728,  C  $546. 

9.  A  and  B  are  partners:  A's  capital  is  $4200;  B's, 
$5600:  after  4  months,  how  much  must  A  put  in,  to  en- 
title him  to  ^  the  year's  gain?  .i»s.  $2100. 

1 0.  A's  capital  is  $9750,  B's  $10500,  C's  $12250 :  after 
3  months,  A  takes  out  $1500,  B  $1250,  C  $1750:  after  4 
months  more,  what  must  A  and  B  put  in,  to  entitle  each 
to  J  of  the  year's  gain?  Ans.  A  $5550,  B  $3300. 

11.  A,  B   and   C    are   in    partnership,  with   capitals  of 
$10800,  $14400,  and  $18000:  after  2  mon.,  A  draws  out 
$600,  in  3  mon.  more  he  draws  out  $1200;  and  in  4  mon. 
more  puts  in  $1000.      B  draws   out   $2400  in    the  first 
6  mon.;  $800    more    in    4  mon.;    and    in    1  mon.    returj»a 


BANKRUPTCY.  321 


$1600.    C  puts  in  $2000  at  the  end  of  5  mon.;  and  4  mon. 
after,  draws  out  $4000.     Divide  their  year's  gain,  §14838. 
Am.  A  $3546;  B  $4752;  C  $6540. 

12.  A,   B  and  C  form  a  partnership,  with   capitals  of 
$10000,  $20000,  $30000,    respectively.      A    draws   out 
$1000  a  year,  B  $1600  a  year,  and  C  $1800  a  year.     In 
5  years,  their  capital  is  $57200:  how  much  of  it  does  each 

wn?  Am.  A  $8000;  B  $18300;  C  $30900. 

13.  A  and  B  enter   into   partnership,  with  $2500  and 
$4000  capital,  respectively.     A   draws  out  $600  after  3 
months,  and  $600  after  6  months  more;   B  draws  $800  at 
each  of  these  times.     At  the  end  of  the  year,  they  have 
$2680:  divide  it.  Am.  A  $920;  B  $1760. 

14.  A  and  B  go  into  partnership,  each  with  $4500.     A 
draws  out  $1500,  and  B  $500,  at  the  end  of  3  mon.,  and 
each  the  same  sum  at  the  end  of  6  and  9  mon. :  at  the  end 
of  1  yr.  they  quit  with  $2200  :  how  must  they  settle  ? 

Ans.  B  takes  $2200,  and  has  a  claim  on  A  for  $300  ? 

15.  A's  gain  is  $1800,  B's  $2250,  C's  $3200;   A's 
capital  was  in  6  mon.;  B's,  9  mon.;  and  C's,  1  year  4  mon.: 
how  much  of  the  capital,  $27450,  did  each  own? 

Ans.  A,  $10980;  B,  $9150;  0,  $7320. 

SUGGESTION. — Divide  each  one's  gain  by  the  number  of  month? 
his  capital  was  used,  to  get  each  one's  gain  for  1  month,  divide  the 
whole  capital  in  proportion  to  these  quotients. 

16.  A,  B,  C  and  D  go  in  partnership:  A  owns  12  shares 
of  the  stock;  B,  8  shares;  C,  7  shares;  D,  3  shares.     After 
3  mon.,  A  sells  2  shares  to  B,  1  to  C,  and  4  to  D;  2  mon. 
afterward,  B  sells  1  share  to  C,  and  2  to  D;  4  mon.  after- 
ward, A  buys  2  from  C  and  2  from  D.     Divide  the  year's 
gain  ($18000.) 

An*.  A  $4650,  B  $4650,  C  $4700,  D  $4000. 


BANKRUPTCY. 

ART.  372.  A  bankrupt,  or  insolvent,  is  one  who,  from 
want  of  means,  is  unable  to  carry  on  his  business. 

The  property  of  a  bankrupt  is  usually  entrusted  to  an 
assignee,  who  converts  it  as  fast  as  possible  into  money,  an* 
pays  the  debts  pro  rata ;  that  is,  so  that  each  creditor  re- 
ceives the  same  part  of  his  claim. 


322  RAPS   HIGHER   ARITHMETIC. 

TO   DIVIDE   THE   PROPERTY   OP   A   BANKRUPT, 

RULE. — Divide  the  whole  property  among  the  creditors  in  pro 
portion  to  their  claims. 

NOTES.  —  1.  All  necessary  expenses,  including  assignee's  fee 
(which  is  genei-ally  a  certain  rate  per  cent,  on  the  whole  amount  of 
property),  must  be  deducted,  before  dividing. 

2.  The  amount  paid  on  a  dollar  can  be  found  by  taking  such  a  part  of 
$1  as  the  whole  amount  of  the  debts  is  of  the  whole  property ;  each  cre- 
ditor's proportion  may  be  then  found  by  multiplying  his  claim  by 
the  amount  paid  on  the  dollar. 

1.  A  has  a  lot  worth  $8000,  good  notes  $2500,  and 
cash  $1500 ;  his  debts  are  $20000 :  what  can  he  pay  on  $1, 
and  what  will  A  receive,  whose  claim  is  $4500? 

Sot. — $8000  -{-  $2500  -}-  $1500  =  $12000,  the  amount  of  property 
which  is  A  §  g  g  g,  or  £  of  the  whole  debts.  Hence,  f  of  $1  =  60  ct.,  the 
amount  paid  on  $1,  and  $-1500  X  .60  =  $2700,  the  sum  paid  to  A. 

2.  My  assets  are  $2520;  I  owe  A  $1200;    B,  $720; 
C,  $600;  D,  $1080:  what  does  each  get,  and  what  is  paid 
on  each  dollar? 

AM.  A  $840;  B  $504 ;  C  $420;  D  $756;  70  ct.  on  $1. 

3.  A  bankrupt's  estate    is  worth   $16000;    his   debts, 
$47500.    The  assignee  charges  5  %.    What  is  paid  on  $1? 
and  what  does  A  get,  whose  claim  is  $3650  ? 

Ans.  32  ct.  on  $1,  and  $1168. 


GENERAL    AVERAGE. 

ART.  373.  General  Average  is  the  method  of  apportioning 
among  the  owners  of  a  ship  and  its  cargo,  any  loss  or  ex- 
pense incurred  during  a  voyage,  for  their  general  benefit, 
such  as  sacrificing  a  part  of  the  cargo  to  save  the  rest,  or 
making  necessary  repairs. 

Such  loss  or  expense  is  borne  by  the  ship,  the  freight,  and 
the  cargo,  in  proportion  to  their  values ;  the  loss  borne  by 
the  cargo  is  divided  among  the  several  shippers  in  propor- 
tion to  the  value  of  their  goods. 

REVI  E  w. — 372.  What  is  a  bankrupt?  What  is  done  with  a  bankrupt1! 
property  ?  What  is  the  rule  for  dividing  a  bankrupt's  property  among  hU 
creditors?  What  must  be  paid  before  distribution?  How  can  the  amount 
paid  on  a  dollar  be  found  ?  How  can  each  creditor's  share  be  then  found  ? 


GENERAL  AVERAGE.  323 


In  estimating  these  contributory  interests,  as  they  are  called,  it  is 
customary  to  value  the  goods  on  board  at  the  price  they  would  bring 
at  the  port  to  which  they  are  bound ;  while  the  freight  is  the  amount 
of  money  received  for  freight  and  pottage,  less  \  for  seamen's  wages, 
(in  New  York,  less  one  half). 

If  the  loss  is  for  repairs,  throw  off  ^  from  the  cost  of  the  new 
masts,  rigging,  &c.,  as  they  are  considered  that  much  better  than 
the  old. 

RULE. —  ifier  ascertaining  the  contributory  interests,  divide  the 
loss  among  them,  in  proportion  to  their  values. 

The  goods  thrown  overboard  (jettison)  are  reckoned  a  part  of  the 
cargo,  and  bear  their  proportion  of  loss. 

1.  The  ship  Dolphin,  overtaken  by  a  storm,  throws  over- 
board goods  worth  $4000,  and  puts  into  Fayal  for  repairs. 
The  necessary  expenses  of  detention  were  $250;  repairs 
$1200.  Divide  the  loss,  estimating  the  vessel  at  $36000, 
the  freight  at  $3750,  the  cargo  at  $52400.  A's  interest  ia 
$7500,  B's,  $16000,  C's,  $10500,  D's,  $12000,  E's, 
$6400.  The  property  lost  was  $2600  of  A's,  $900  of 
B's,  and  $500  of  C's. 

SOLUTION. 

CONTRIBUTORY    INTERESTS.  DAMAGES. 

Vessel $36000          Goods  lost,     ....  $4000 

Freight  (less  ^),  .     .       2500          Expense  of  detention,        250 
Cargo, 62400          Repairs  (less  \),    .    .       800 

Total,    ....  $90900  Total  loss,    .     .    .  $5050 

Then  /oVoo  =  jVi  tno  Part  tnat  eacn  interest  loses. 

The  owners  of  the  vessel  and  freight  pay  $38500  X  yg  =  $2138| 

The  owners  of  the  cargo  pay  $52400  X  y  5    •    •     •    •    =2911$ 

Total  loss, $5050 

Owners  of  the  vessel  pay  $2138| ,  and  re-          «CBIVED.       PAID. 

ceive  $1050,  or  pay 1088§ 

A  pays  $41G|  and  receives  $2600,  or  receives  $2183^, 
B  pays  $888§  and  receives  $900,  or  receives         11$, 

C  pays  $583  3  and  receives  $500,  or  pays 83  i 

D  pays 666| 

E  pays _. ._  .    .  355$ 

2194^,        2194~! 

REVIEW  . — 373.  What  is  general  average  ?  How  is  loss  at  sea  borne  ? 
Which  are  the  contributory  interests?  How  are  the  goods  valued?  How 
IB  the  freight  valued?  AVhat  deduction  is  made  from  the  cost  of  repairs? 
After  the  contributory  interests  have  been  estimated,  how  is  the  loss  dU 
rirted  ?  Why  should  the  goods  lost  bear  their  proportion  of  the  loss  ? 


324  RAY'S   HIGHER   ARITHMETIC. 


-.  The  brig  A  Jams,  ni'  N.>\v  York,  bound  to  New  Orleans, 
suffered  damage,  $480,  and  loss  of  cargo,  §5600.  The 
vessel  was  valued  at  $18800;  the  freight,  $3200;  the 
cargo,  $29600.  Divide  the  loss;  and  settle  A's  account, 
who  shipped  $14400,  and  lost  $1800. 

Am.  Kate  of  loss  cVs  ;  ship  pays  $2415.36,  and  receives 
$320,  making  a  balance  paid,  $2095.36 ;  cargo  paya 
$3504.64;  A  pays  $1704.96,  and  receives  $1800,  mak 
ing  a  balance  received  of  $95.04 

3.  The  schooner  Washington,  crippled  by  a  storm,  was 
relieved  by  the  ship  Leopard.  The  repairs  cost  $1350, 
salvage  $4500.  The  vessel  is  worth  $22000;  the  freight, 
$3450.  The  cargo  was  owned  by  A,  B,  C,  and  D  :  A, 
$10800  ;  B,  $16200  ;  C,  $19652  ;  D,  $7348.  What  is 
the  rate  of  loss,  and  what  does  each  pay  ? 

Ans.  Js  rate  of  loss  ;  A  pays  $744.83  ;  B,  $1117.24 ;  C, 
$1355 .31 ;  D,  $506 . 76 ;  ship  pays  $1675 . 86,  and  receive? 
$900,  making  a  balance  to  be  paid  of  $775.86 


RATE  BILLS  FOR  SCHOOLS. 

ART.  374.  Another  application  of  proportional  parts  is 
in  making  out  rate-bills  for  public  schools,  in  districts  where 
they  are  not  supported  by  general  tax. 

RULE. — From  the  lohole  expenses  of  the  school  deduct  the  public 
money,  if  any,  and  divide  the  remainder  among  the  families  of  the 
district,  in  proportion  to  the  number  of  days  of  attendance  of  each. 

NOTE. — It  is  generally  more  convenient  to  find  first  the  rate  per 
day.  by  dividing  the  expense  to  be  distributed  by  the  whole  number 
of  days  of  attendance,  and  then  make  out  each  pupil's  bill,  by  multi- 
plying his  number  of  days  of  attendance  by  this  rate. 

1.  A  school  pays  $500  for  teacher's  salary,  $26.50  for 
repairs,  and  $12.30  for  fuel,  and  draws  $50  public  money: 
the  whole  number  of  days  of  attendance  is  3640.  Find  the 
ate  per  day,  and  the  bill  of  A,  who  sends  one  pupil  175 
lays ;  of  B,  who  sends  2  pupils,  220  days  each  :  of  C,  who 
Bends  one  pupil  108  days,  one  76  days,  and  one  192  days. 

Ans.  Rate  13|  ct.;  A  pays  $23. 50;  B  $59. 09;  C$50.49 


R  E  v  i  EW. — 374.  What  is  the  rule  for  making  out  rate  bills  for  schools  1 
What  is  convenient  in  applying  the  rule  ? 


ALLIGATION.  325 


2.  The  salary  is  $180 ;  other  expenses,  $18.75 ;  the  public 
money,  $30;  the  whole  number  of  days  of  attendance,  2562: 
what  is  the  rate  per  day?  and  A's  bill,  who  sends  for  87  da.? 

Ans.   6g54-ct.  rate;  A  pays  $5.73 

3.  The  expenses  are  $312  ;  the  whole  number  of  days  of 
attendance,  4160  :  what  is  the  rate  per  day?  and  D's  bill, 
who  sends  1  pupil  116  days,  and  another  98  days? 

Ans.  Rate  7s  ct.;  D's  bill  $16.05 

4.  The  salary  is  $35  per  month  ;  the  other  expenses  for 
1  quarter,  $6.80;  the  public  money,  $10.20;  the  whole 
number  of  days  of  attendance,  1120  :  what  is  the  rate  per 
day?  and  E's  bill,  who  sends  48  days? 

Ans.  Rate  9 Act.;  E's  bill  $4.35 


XXXII.   ALLIGATION. 

ART.  375.  Alligation  is  a  method  of  finding  the  value 
per  lb.,  bu.,  &c.,  of  a  mixture,  when  the  quantity  and  price 
of  its  several  ingredients  are  known. 

It  is  also  used  in  getting  the  mean  or  average  result  of  several 
quantities  of  the  same  kind,  but  differing  in  degree,  such  as  obser- 
vations by  instruments,  measurements  of  irregular  figures,  &c. 

The  price  or  other  result  obtained  by  Alligation  is  called  the 
mean  or  average. 

CASE  I. — To  find  the  ayerage  price  of  a  mixture,  when 
the  quantity  and  cost  of  each  ingredient  are  known. 

RULE. — Find  the  value  of  each  ingredient  at  its  price;  add  these 
values  for  the  value  of  the  mixture;  divide  the  sum  by  the  sum  of 
the  quantities  of  the  ingredients. 

NOTE. — Express  each  ingredient  in  the  same  denomination. 

If  3  lb.  of  sugar  at  5  ct.  a  lb.  and  2  lb.  at  4|  ct.  a  lb.  be 
mixed  with  9  lb.  at  6ct.  a  lb.,  what  per  lb.  is  the  mixture 
worth  ? 

SOLUTION.— The  3  lb.  at  5  ct.  per  lb.  =  15  ct.;  the  2  lb.  at  4^  ct. 
per  lb.  =  9  ct. ;  the  9  lb.  at  6  ct.  per  lb.  =  54  ct. :  therefore,  the 
whole  14  lb.  are  worth  78  ct.,  =  78  -f-  14  =  5$  ct.  per  lb.  Ans. 

1.  Find  the  average  price  of  6  lb.  tea  at  80  ct.,  15  lb.  at 
50  ct.,  5  lb.  at  60  ct.,  9  lb.  at  40  ct.  Ans.  54ct.  per  lb. 

REVIEW. — 375.  What  is  Alligation?  What  is  the  result  called? 
What  is  Case  1?  The  rule?  How  must  the  quantity  of  each  ingredient 
be  expressed  ?  Solve  the  example. 


826  RAY'S   HIGHER  ARITHMETIC. 

2.  The  average  price  of  40  hogs  at  §8  each,  30  at  §10  each, 
16  at  812.50  each,  54  at  $11.75  each.        Ans.  §10.39  each. 

3.  How  fine  is  a  mixture  of  5  pwt.  of  gold,  10  carats  fine ;  2 
pwt.  18  carats  fine;  6  pwt.,  20  carats  fine,  and  1  pwt  pure  gold? 

Ans.  18?  carats  fine. 

4.  Find  the  specific  gravity  of  a  compound  of  15  Ib.  of  copper, 
specific  gravity,  7| ;  8  Ib.  of  zinc,  specific  gravity,  G|  j  and  \  Ib. 
M"  silver,  specific  gravity,  10^.  Ans.  7.445 — 

EXPLANATION. — The  Specific  Gravity  of  a  body  is  ita  own   weight  divided  by  th* 
eight  of  an  equal  bulk  of  water. 

5.  What  per  cent,  of  alcohol  in  a  mixture  of  9  gal.,  80  Jo 
strong;  12  gaL,  92  %  strong;   10  gal.,  95  %  strong;  and  11  gal., 
98  #  strong?  Am.  93  %. 

ART.  376a.  CASE  n. 

To  find  what  proportions  of  several  ingredients,  whose  prices 
are  known,  must  be  used,  to  make  a  mixture  of  a  given  price. 

1.    What  relative  quantities  of  sugar  at  9  ct.  a  Ib.  and  5  ct.  a 
Ib.  must  be  used  for  a  compound  at  6  ct.  a  Ib.  ? 

ANALYSIS. — If  you  put  1  Ib.  OPERATION. 


at  9  ct.  in  the  mixture  to  be  sold 
for  6  ct.,  you  lose  3  ct. ;  if  you 
put  1  Ib.  at  5  ct.  in  the  mixture 


5.   .    .  3  Ib.  at  5  ct.  =  15  ct. 
9    .    .  1  Ib.  at  9  ct.  =    9  ct. 


to  be  sold  at  6  ct.,  you  gain  1  ct.;  4  ^.      worth      24  ct. 

3  such  Ib.  gain  3  ct:  the  gain  and         which  is  2j4  =  G  ct.  a  Ib. 
loss  would  then  be  equal  if  3  Ib.  at  5  ct.  are  mixed  with  1  Ib.  at  9  ct 

RULE. — Place  the  prices  of  the  ingredients  in  a  vertical  row  in 
regular  order,  having  the  smallest  at  the  top,  and  the  largest  at  the 
bottom.  To  the  left  of  this  row  draw  a  vertical  line,  and  on  the 
other  side  of  it  set  the  price  of  the  mixture  o})posite  the  place  it 
would  occupy,  if  it  stood  in  the  right-hand  column  of  prices. 

Then  connect  by  a  curved  line  any  two  numbers  in  the  riglit-liand 
column,  one  of  which  is  greater  and  one  less  than  the  mean  price ; 
when  each  of  the  right-hand  numbers  has  been  thus  connected  with 
another,  take  the  difference  between  each  of  them  and  the  mean  price 
on  the  left,  and  set  this  difference  opposite  the  number  with  which 
it  is  connected. 

After  all  the  differences  have  been  taken,  the  proportional  quan- 
tity at  each  price  will  be  the  number  standing  opposite  that  price, 
unless  there  be  several  such  numbers,  in  which  case,  their  sum  will 
be  the  proportional  quantity  at  that  price. 

NOTE. — One  number  may  be  connected  with  two  or  more  others; 
if  BO,  several  numbers  will  stand  opposite  it,  and  their  sum  must  be 
taken  for  its  proportional. 

PROOF. — With  the  proportional  quantities  thus  found,  determine 


ALLIGATION.  327 


by  Case  I  the  mean  price  of  the  mixture ;  if  it  agrees  with  the 
mean  price  given,  the  work  is  right. 

2.  What  relative  quantities  of  tea  worth  25,  27,  30,  32,  and 
45  ct.  per  Ib.  must  be  taken  for  a  mixture  worth  28  ct.  per  lb.? 

Ans.  19,  4,  3,  1,3  lb.  respectively. 

REM. — It  is  evident  that  other  results  may  be  obtained  by  making 
the  connections  differently;  as  6,  17,  3,  3,  1  lb.,  or  17,  6,  1,  1,  3  lb. 

3.  What  of  sugar  at  5,  5j,  6,  7,  and  8  ct.  per  lb.  must  be 
taken  for  a  mixture  worth  6|  ct.  per  lb.  ? 

Ans.  1,  5,  5,  7,  8  lb.  respectively;  or,  5,  1,  1,  8,  7  lb.,  <tc. 

SUGGESTION. — When  the  proportional  quantities  determined  by 
the  rule,  contain  fractions,  multiply  them  all  by  the  least  common 
multiple  of  the  denominators,  which  converts  them  into  whole  num- 
bers having  the  same  relative  values. 

4.  What  relative  quantities  of  alcohol,  84,  86,  88,  94  and 
96  %  strong,  must  be  taken  for  a  mixture  87  %  strong  ? 

Ans.  10,  7,  3,  1,  3  gal. ;  or,  7,  10,  1,  3,  1  gal.,  &c. 

5.  What  of  gold  and  silver,  whose  specific  gravities  are  19^ 
and  10^,  will  make  a  compound  whose  specific  gravity  shall  be 
16.84?  Ans.  723  lb.  silver  to  3487  'lb.  gold. 

6.  What  of  silver  |  pure,  and  fj  pure,  will  make  a  mixture  | 
pure?  -4ns.  1  lb.,  |  pure  ;  5  lb.,  jJo  pure. 

7.  What  of  pure  gold  (24  carats),  and  18  carats,  and  20  carats 
fine,  must  be  taken  to  make  22  carat  gold  ? 

Ans.  1  part  18  carats,  1  part  20  carats,  3  pure. 

Suo. — Any  common  factor  may  be  omitted  from  the  proportionals. 

CASE  III. 

ART.  376".  Given,  the  prices  of  the  ingredients,  and  the  quan 
tity  and  price  of  the  mixture,  to  find  the  quantity  of  each  ingre 
dient 

RULE. — Find  the  relative  quantities  of  the  ingredients  according 
to  the  last  rule;  divide  the  given  quantity  of  the  mixture  into 
parts  proportional  to  these  numbers,  by  rule  in  Art.  369 :  these 
results  will  be  the  quantities  of  the  ingredients  required. 

PROOF. — Add  the  quantities  of  the  ingredients ;  also  add  the 
cost  of  the  ingredients  at  the  given  prices :  these  sums  must  agree 
with  the  quantity  and  cost  of  the  mixture. 

How  many  bushels  of  peaches  at  20,  30,  37,  40,  and  50  ct.  a 
bu.,  will  make  a  lot  of  58  bu.  3  pk.  at  35  ct  a  bushel  ? 


REVIEW. — 376a.  What  is  Case  2?  Solve  the  example.  What  is  the 
rule?  The  proof?  When  will  it  be  necessary  to  connect  one  price  with 
kwo  of  th»  others  ?  What  is  Case  3  ? 


328  RAY'S   HIGHER   ARITHMETIC. 


SOLUTION.  -By  last  case  the  proportional  quantities  of  the  ingre- 
dients are  2,  HO,  15,  5  and  5.  Dividing  58  bu.  :>>  pk.  or  235  pk.  in  pro- 
portion to  these  numbers,  by  rule  in  Art.  3G9,  gives  2bu.  2pk.;  25 bu., 
18bu.  3  pk. ;  6  bu.  1  pk. ;  G  bu.  1  pk.,  respectively,  at  the  prices  men- 
tioned. 

1.  How  much  copper,  specific  gravity  7f ,  with  silver,  specific 
gravity  1(H,  will  make  lib.  Tr.,  of  spe.  grav.  8|? 

Ans.  7|f  I  ox.  copper,  4||§  oz.  silver. 

2.  How  much  gold  15  carats  fine,  20  carats  fine,  and  pure,  will 
make  a  ring  18  carats  fine,  weighing  4  pwt.  16  gr.  ? 

Ans.  2  pwt.  16  gr.  ;   1  pwt. ;   1  pwt. 

3.  Hiero,  king  of  Syracuse,  gave  his  goldsmith  14  Ib.  of  gold 
and  3w  Ib.  of  silver  to  make  a  crown :  suspecting  that  the  gold 
had  not  been  all  used,  he  requested  Archimedes  to  find  how  much 
had  been  abstracted,  the  specific  gravity  of  gold  being  19 1  ;  of 
silver  1(U;  and  of  the  crown,  14f. 

Ans.  It  contained  10!j|lb.  of  gold  and  6f£  Ib.  of  silver;  3^1b 
of  gold  had  been  replaced  by  silver. 

CASE  IV. 

ART.  377.  Given,  the  price  of  the  mixture,  the  prices  of  the 
ingredients,  and  the  quantity  of  one  ingredient,  to  find  the  quan- 
tities of  the  other  ingredients,  and  of  the  mixture. 

RULE. — Find  the  relative  quantities  of  the  ingredients,  by  Casel; 
the  quantity  of  each  ingredient  will  be  such  a  part  of  the  given 
quantity,  as  its  proportional  is  of  the  proportional  belonging  to 
the  given  quantity. 

PROOF. — By  Case  I. 

NOTE. — After  the  quantities  of  all  the  ingredients  have  been 
found,  their  sum  will  be  the  quantity  of  the  mixture. 

How  many  bushels  of  hops,  worth  respectively  50,  60  and  75  ct. 
per  bu.,  with  100  bu.  at  40  ct.  per  bu.,  will  make  a  mixture  worth 
65  ct.  a  bu.  ? 

SOLUTION. — By  Case  II  the  proportionals  are  10, 10  and  45  of  the 
first  three  sorts  to  10  of  the  last ;  hence,  the  quantities  of  the  two 
first  sorts  must  be  j$  of  100  bu.  =  100  bu. ;  the  quantity  of  the  third 
§ort  is  f  g  of  100  bu.  =  450  bu. 

1.  How  many  railroad  shares  at  50^  must  A  buy,  who  has  80 
shares  that  cost  him  72$,  in  order  to  reduce  his  average  to  60$? 

Ans.  96  shares. 

2.  I  bought  2000  cwt  of  pork  at   85.80  a  cwt  :    how  much 
must  I  buy  at  $4.75  a  cwt,  so  as  to  average  85. '25  per  cwt.  ? 

Ans.  2200  cwt. 


REVIEW.— 3766.  What  is  the  rule?    The  proof?     ?,77.  What  is  Case  4! 
The  rule?    The  proof? 


ALLIGATION.  329 


3.  A  jeweller  has  3  pwt.  9  gr.  of  old  gold,  16  carats  tine  ;  how 
much  U.  S.  gold,  21 1  carats  fine,  must  he  mix  with  it,  to  make  it 
18  carats  tine  ?  Ans.  1  pwt.  21  gr. 

4.  How  much  water  (0  per  cent.)  will  dilute  3  gal.  2qt.  1  pt. 
of  acid  .)!  %  strong,  to  56$  ?  Ans.  2  gal.  1  qt.  $  pt. 

5.  1  mixed  1  gal.  2  qt.  A  pt.  of  water  with  3  qt.  lA  pt.  of  pure 
acid  :  the  mixture  has  \byc  more  acid  than  desired:  how  much 
water  will  reduce  it  to  the  required  strength  ? 

Ana.  1  gal.  2  qt.  1^  pt. 

CASE  V. 

ART.  378.  Given,  the  price  of  the  mixture,  the  price  of  each 
ingredient,  and  the  quantities  of  two  or  more  ingredients,  to  find 
the  quantities  of  the  remaining  ingredients  and  of  the  mixture. 

RULE. — Find  by  Case  I,  the  quantity  and  average  price  of  a 
mixture  composed  of  those  ingredients  whose  quantities  and  prices 
are  known ;  consider  these  an  the  quantity  and  price  of  a  single 
ingredient,  and  find  the  quantities  of  the  remaininy  ingredients, 
and  of  the  whole  mixture,  by  Case  I  T. 

PROOF. — By  Case  I. 

A  buys  400  bbl.  of  flour  at  $7.50  each,  640  bbl.  at  $7.25,  and 
960  bbl.  at  86.75:  how  many  must  he  buy  at  §5.50,  to  reduce 
his  average  to  §6.50  per  bbl.  ? 

SOLUTION. — The  mean  price  of  the  first  three  items,  by  Case  I,  is 
§7.00  a  piece  for  2000  bbl.  Then  with  $7.00,  and  $5.60  for  the  prices 
of  the  ingredients,  and  $0.50  for  the  price  of  the  mixture,  and 
2000  bbl.  for  the  quantity  of  the  first  ingredient,  find,  by  Case  IV, 
the  quantity  of  the  other  ingredient,  ==  1120  bbl. 

1.  How  much  lead,  specific  gravity  11,  with  4  oz.  copper,  sp. 
gr.  9,  can  be  put  on  12  oz.  of  cork,  sp.  gr.  \,  so  that  the  3  will 
|ust  float,  that  is,  have  a  sp.  gr.  (1)  the  same  as  water? 

Ans.  2  Ib.  7£  oz. 

2.  How  much  water,  with  3  pt.  of  alcohol,  96  fo   strong,  and 
8  pt.,  78  %,  will  make  a  mixture  60  fo  strong?        Ans.  4^  pt. 

3.  How  many  shares  of  stock  at  40  %  must  A  buy,  who  has 
bought  120  shares  at  74  fc,  150  shares  at  GSfo,  and  '130  shares 
at  54  %    so  that  he  may  sell  the  whole  at  60  fc,  and  gain  20  %  ? 

Ans.  610  shares. 
CASE   VI. 

ART.  379.  Given,  the  quantity  and  price  of  the  mixture,  the 
quantities  and  prices  of  one  or  more  ingredients,  and  the  prices  of 
the  remaining  ingredients,  to  find  the  quantities  of  the  remaining 
ingredients. 

R  E v  i  KW.—378.  What  is  Cnao  5  ?     Tho  rule  ?     The  proof? 
2* 


RAY'S   HIGHER   ARITHMETIC. 


RULE. — From  the  quantity  and  cost  of  the  whole  mixture,  deduct 
the  quantities  and  costs  of  the  given  ingredients:  the  remainders 
will  be  the  quantity  and  cost  of  a  mixture  composed  of  the  remain- 
ing ingredients :  from  which,  the  quantities  of  those  ingredient* 
can  be  found  by  Case  III. 

PROOF. — By  Case  I. 

What  quantities  of  sugar  at  3  ct.  per  Ib.  and  7  ct.  per  Ib.  ;  with 
2  Ib.  at  Sec.,  and  51b.  at  4  ct.  per  Ib.,  will  make  16  Ib.  worth 
Get.  per  Ib.? 

S  o  L. — The  2  Ib.  at  8  ct.  and  the  5  Ib.  at  4  ct.,  make  a  mixture  of 
7  Ib.  worth  36  ct.,  which,  deducted  from  the  16  Ib.  worth  96  ct.,  leaves 
9  Ib.  worth  60  ct.,  or  9  Ib.  at  Of  ct.  a  Ib.  Taking  9  Ib.  and  6  jj  ct.  as  the 
quantity  and  price  of  the  mixture  whose  ingredients  are  worth  3  ct. 
and  7  ct.  a  Ib.,  find,  by  Case  III,  the  quantities  of  the  ingredients, 
$  Ib.  at  3  ct.,  and  8}  Ib.  at  7  ct.  Ans. 

REMARK. — There  must  be  at  least  two  ingredients,  whose  quanti- 
ties are  required. 

1.  How  many  bbl.  flour  at  §8,  and  $8.50  ;  with  300  bbl.  at 
§7.50,  and  800  at  §7.80,  and  400  at  §7.65,  will  make  2000  bbl. 
at  §7.85  a  bbl.  ?  Ans.  200  bbl.  at  §8  ;  300  bbl.  at  §8.50 

2.  What  quantities  of  tea  at  25  ct.,  and  35  ct,  a  Ib. ;  with  14  Ib. 
at  30  ct.,  and  20  Ib.  at  50  ct.,  and  6  Ib.  at  60  ct.,  will  make  56  Ib 
at  40  ct.  a  Ib.  ?  Ans.  10  Ib.  at  25  ct.,  and  6  Ib.  at  35  ct 


XXXIII.    INVOLUTION. 

ART.  380.    INVOLUTION  is  the  process  of  finding  a  power. 

A  power,  is  the  product  of  a  number  by  itself  one  or  more  times. 

The  number  from  which  the  powers  arise,  is  called  the  root  of 
those  powers,  and  sometimes  the  firsi  power. 

Powers  are  of  different  degrees,  2d,  3d,  4th,  &c.,  according  to 
the  number  of  times  the  root  is  used  in  their  formation. 

The  degree  of  a  power  is  indicated  by  an  exponent,  which  is  a 
email  figure,  placed  to  the  right  of  the  root;  thus,  7~  signifies 
the  2d  power  of  7;  53,  the  third  power  of  5. 

The  2d  power  of  a  number  is  called  its  square,  because  the  area 
Oi  a  square  is  obtained  by  forming  a  2d  power;  viz..  the  product  of 
the  number  of  linear  units  in  one  side  by  itself. 

The  3d  power  of  a  number,  is  called  its  cube,  because  the  solidity 


REVIEW.  —  379.   What  is  Case  6?     Tho  rule?     Tho  proof?     How 
many  ingredients  must  there  be,  whoso  quantities  are  required  ? 


INVOLUTION.  33 1 


of  a  cube  is  obtained  by  forming  a  3d  power;  viz.,  the  product  of  the 
number  of  linear  units  in  one  side  by  itself  twice. 

TO    FIND   ANY    POWER   OP   A   NUMBER. 

ART.  381.  RULE. — Multiply  the  number  continually  by  itself, 
until  it  has  been  used  as  often,  as  the  degree  of  the  j>oiver  indi- 
cates :  the  last  product  will  be  tJie  power  required. 

NOTES. — 1.  The  number  of  multiplications  will  be  one  less  than  the 
exponent,  because  the  root  is  used  twice  in  the  first  multiplication, 
once  as  multiplicand  and  once  as  multiplier. 

2.  When  the  power  to  be  obtained  is  of  a  high  degree,  multiply  by 
some  of  the  powers  instead  of  by  the  root  continually;  thus,  to  get 
the  9th  power  of  2,  multiply  its  6th  power  (64)  by  its  3d  power  (8); 
or,  its  5th  power  (32)  by  its  4th  power  (16):  the  rule  being,  that  tht 
product  of  any  two  powers  of  a  number  is  that  power  whose  degree  is 
equal  to  the  sum  of  their  degrees. 

3.  Any  power  of  1  is  1;  any  power  of  a  number  greater  than  1  is 
greater  than  the  number  itself:  any  power  of  a  number  less  than  1, 
is  less  than  the  number  itself. 

ART.  382.  From  Note  2,  last  Art,  43  X  43  X  43  X  43  X 
43  =  41  6,  but  the  expression  on  the  left  is  the  5th  power  of  43: 
hence,  (43)°  =  41  5 ;  that  is,  when  the  exponent  of  the  power  re- 
quired is  a  composite  number  (15),  raise  the  root  to  a  power  whose 
exponent  is  one  of  its  factors  (3),  and  this  result  to  a  power  whose 
exponent  is  the  other  factor  (5). 

ART.  383.  Any  power  of  a  fraction  is  equal  to  that  powei 
of  the  numerator  divided  by  that  power  of  the  denominator. 

To  raise  a  mixed  number  to  any  power,  reduce  it  to  a  fractional 
form,  and  then  proceed  as  just  directed. 

ART.  384.  The  square  of  a  decimal  must  contain  twice,  it* 
cube,  three  times  as  many  decimal  places  as  the  root,  &c.:  hence, 
to  obtain  any  power  of  a  decimal,  proceed  as  if  if  were  a  wholf 
number,  and  point  off  in  the  result  a  number  of  decimal  placet 
equal  to  the  number  in  the  root  multiplied  by  the  exponent  (J 
the  poicer. 

EXAMPLES  FOR  PRACTICE. 

1.  (5)'   . 

2.  143    . 

3.  66      , 

4.  192* 

5.  I10    . 

e.  m4  . 

7. 


ANS. 

ANS. 

.  =25. 

8. 

(I)5 

1  r,n  n7 
.   .  =  32758 

=  2744. 

9. 

(.02)3 

.  =.000008 

=  7776. 

10. 

(54)2  = 

58,  or  390625 

=  36864. 

11. 

(.046)3 

=  .000097336 

.   =1.J12. 

,  ,x7 

i 

.  =  jj8215  i  13. 

2056-  . 

.  =4227136. 

.=llif  |14. 

(7.622) 

1  .  =  58.1406.} 

332  RATS   HIGHER   ARITHMETIC. 


XXXIV.    EVOLUTION. 

ART.  385.    Evolution  is  the  process  of  finding  a  root. 

A  root  of  a  number  ia  another  number,  of  which  the  given 
number  is  some  power. 

Evolution  is  the  reverse  of  Involution,  and  is  sometimes  called 
the  Extraction  ofroota. 

ART.  386.  Roots,  like  powers,  are  divided  into  degrees,  3d, 
3d,  4th,  &c.:  the  degree  of  a  root  is  always  the  same  as  the  de- 
gree of  the  power,  to  which  that  root  must  be  raised,  to  produce 
the  given  number. 

Thus,  the  3d  root  of  343  is  7,  since  7  must  be  raised  to  the  3d 
power,  to  produce  343:  the  5th  root  of  1024  is  4,  since  4  must  be 
raised  to  the  5th  power,  to  produce  1024. 

Since  the  '2d  nud  3d  powers  are  called  the  square  and  cube,  so  the 
2d  and  3d  roots  are  called  the  square  root  and  cube  root. 

ART.  387.  To  indicate  the  root  of  a  number,  use  the  radical 
sii/n  (>J),  of  fractional  exponent. 

The  radical  sign  is  placed  before  the  number ;  the  degree  of 
the  root  is  shown  by  the  small  figure  between  the  branches  uf 
the  radical  sign,  called  the  index  of  the  root. 

Thus,  3/18  signifies  the  cube  root  of  18;  ^/9  signifies  the  5th  root 
of  9.  The  square  root  is  usually  indicated  without  the  index  2; 
thus,  ^/10  is  the  same  as  ^/10. 

The  root  of  a  number  may  be  expressed  by  a  fractional  expo- 
nent whose  numerator  is  1,  and  denominator  is  the  index  of  the 
root  to  be  expressed. 

Thus,   Jl  =  7$,  and   is/5  =  6$,  and  t/64  =  64i,  and  so  on. 

NOTE.— r-Any  root  of  1  is  1;  any  root  of  a  number  greater  than  1 
i»  less  than  the  number  itself:  any  root  of  a  number  less  than  1  is 
greater  than  the  number  itself. 


THE    SQUARE    ROOT. 

ART.  388.  The  square  root  of  a  number  is  another  number 
which,  multiplied  by  itself,  will  produce  the  given  number;  thus, 
7  is  the  square  root  of  49,  because  7X7  =  49. 

The  square  root  of  a  number  of  two  figures  is  found  by  trial  and 
proof:  thus  V 04  =  8,  because  8  X  8  =  64.  If  the  root  can  not  be 
pxactly  obtained,  the  number  is  called  an  imperfect  square,  and  ita 
root  is  obtained  by  trial  to  within  unity:  thus,  75  is  an  imperfect 
square,  because,  as  it  lies  between  G4  and  81,  whose  exact  square 


SQUARE  ROOT.  3JJ3 

roots  are  8  :iud  9,  its  square  root  must  be  greater  than  8  and  lesa 
than  9,  either  of  which  is  its  square  root  to  within  unity,  but  neither 
its  square  root  exactly. 

ART.  389.  A  number  of  more  than  two  figures  has  two 
figures  in  its  square  root ;  for,  being  equal  to  or  greater  than  ]  00, 
its  square  root  must  be  equal  to  or  greater  than  10.  To  find  the 
square  root  of  such,  first  prove  this 

PROPOSITION. 

If  a  number  is  composed  of  tens  and  units,  its  square  will  con- 
sist of  the  square  of  those  tens,  with  twice  the  product  of  the  tens 
by  the  units,  and  the  square  of  the  units. 

DEM. — Take  any  number  com-  40  +     7     =     47 

posed  of  tens  and  units,  as  47,  40+     7     =     47 

and  square  it,  first  separating  it  280  +  49          329 

into   its   tens   and   units.      The      1600  +  280  188 

work  shows  that  the  square  of      j  000  +  .560  +  49  =  2209 
17  consists  of  1600,  (the  square 

of  the  tens  40),  of  560,  (twice  the  product  of  the  tens  by  the  units,  2  X 
10  X  7),  aud  of  49,  (the  square  of  the  units):  the  same  is  true  of  any 
other  number  containing  tens  and  units. 

Extract  the  Square  Root  of  7396. 

ANALYSIS. — Since   the   number   7396   has-  7396(86 

more  than  2  figures,  its  root  will  be  composed  64 

of  tens  and  units,  and  the  square  will  be  made       166^996 
up  of  the  square  of  these  tens,  with  twice  the  pro-  996 

duct  of  the  tens  and  units,  and  the  square  of  the 

units.  The  square  of  tens  is  always  hundreds,  and  since  73,  the 
hundreds  of  the  number,  lies  between  64  and  81,  its  square  root  must 
lie  between  theirs,  being  greater  than  8  and  less  than  9 ;  hence,  8  is 
the  tens'  figure  of  the  root,  9  being  too  large.  As  the  two  right-hand 
figures  (96)  are  not  used  in  finding  the  tens'  figure  of  the  root,  point 
them  off  so  as  to  show  distinctly  the  73,  the  only  figures  to  be  ex- 
amined for  that  purpose. 

Subtract  the  square  of  the  tens,  64  (hundreds),  from  the  given  num  • 
her  7396,  which  is  the  square  of  the  tens,  with  twice  the  product  of  the 
tens  by  the  units,  and  the  square  of  the  units  :  the  remainder  996,  must 
be  twice  the  product  of  the  tens  by  the  units,  and  the  square  of  the  units. 

The  square  of  the  units  being  comparatively  small,  may  be 
neglected  for  the  present,  and  the  996  regarded  as  twice  the  product 
of  the  tens  by  the  units:  divide  jt  by  twice  the  tens  (16  tens),  to  get 
the  units.  Dividing  996,  by  160  (16  tens),  or,  what  is  more  convenient, 
lividing  99  by  16,  the  units'  figure  is  found  to  be  6. 


RAY'S  HIGHER   ARITHMETIC. 


To  prove  this  figure  correct,  form- with  it  not  only  twice  the  product 
of  (lie  tens  by  the  units,  but  also  the  square  of  the.  units,  since  the  996 
consists  of  both  these  parts.     To  do  this  conveniently,  set  the  units' 
figure  (0)  on  the  right  of  the  trial  divisor  (16),  and  multiply  the  com- 
plete divisor  (160)  by  the  units'  figure  (6),  thereby  making  twice  the 
product  of  the  tens  by  the  units  (1GO  X  0),  and  the  square  of  the  units 
GX6);  since  this  produces  996  exactly,  86  must  be  the  exact  square 
oot  of  7936 :  the  same  process  and  explanation  apply  in  obtaining 
ihe  square  root  of  any  number,  however  large.     Hence, 

ART.  390.   TO  EXTRACT  THE  SQUARE  ROOT  OP  A  NUMBER, 

RULE. — 1.  Separate  the  given  number  into  periods  of  two  figures, 
commencing  at  units;  the  left  hand  period  may  have  1  or  2  figures. 

2.  Take  the  square   root  of  the  nearest  square  below  the  left 
hand  period :  this  will  be  the  first  figure  of  the  root. 

3.  Subtract  the  square  of  this  figure  from  the  left  hand  period, 
bring  down  the  next  period;    divide  the  result  exclusive  of  Hit 
right  hand  figure,  by  twice  the  part  of  the  root  already  found, 
the  quotient  will  be  the  2d  figure  of  the  root. 

4.  Set  this  figure  of  the  root  on  the  right  of  the  divisor ;  mul- 
tiply the  divisor  thus  completed,  by  the  2d  figure  of  ihe  root;  sub- 
tract the  product  from  the  last  dividend,  and  bring  down  anothei 
period. 

5.  Double  the  root  already  found  for  a  trial  divisor,  find  anothei 
figure  of  the  root,  and  proceed  as  before,  until  all  the  period* 
have  been  brought  down. 

NOTES. — 1.  If  any  product  is  larger  than  the  dividend  from  which 
it  is  to  be  taken,  the  last  figure  of  the  root  is  too  large. 

2.  If  any  dividend,  exclusive  of  its  right-hand  figure,  is  not  large 
enough  to  contain  its  trial  divisor,  place  a  cipher  in  the  root,  and  at 
the  right  of  the  divisor ;  bring  down  another  period  and  continue 
as  before. 

3.  When  a  remainder  is  greater  than  the  previous  divisor,  it  does 
not  follow  that  the  last  figure  of  the  root  is  too  small,  unless  thai 
remainder  is  large  enough  to  contain  twice  the  part  of  the  root  already 
found,  and  1  more,  for  this  would  be  the  proper  divisor  to  go  into  the 
remainder,  if  the  root  were  increased  by  1. 

EXAMPLES  FOR  PRACTICE. 

AN8.  AXS. 

4.  ^57600"      .     .     =  240. 

5.  Jl  6499844      .  =4062. 

6.  ^49098049      .  =7007. 


1.  V2809      .     .     .     =53. 

2.  V1444~   .     .     .     =38. 
3    7TT88T    ...  =109. 


SQUARE   ROOT.  335 


7.  V185C40G25  .  =  13625. 

11.  V73005       .     =270+ 

8.  V80012304    .  =  8945— 

12.  ^3863    .     .  =  7584— 

9.  VG2U3794      .  =  2491— 

13.  V  1245  X  (252)  -=8892- 

10.  ^3444736   .      =1856. 

14.  V(96059601)£      =99. 

15.  V  (126)*  x  (58)*  X  ( 

G04)2     .     .     =4414032. 

ART.  391.     The  square  root  of  a  common  fraction  is  the  J? 
oot  of  its  numerator,  divided  by  the  sq.  root  of  i(s  denominator. 

NOTES. — 1.  Reduce  the  fraction  to  its  lowest  terms  before  com 
mencing  the  operation,  and  if  the  denominator  be  not  a  square,  make 
it  so,  by  multiplying  both  terms  by  the  denominator,  or  some  smaller 
number  that  will  answer  the  purpose. 

2.  The  square  root  of  a  mixed  number  may  be  found  by  first  con- 
verting it  into  a  common  fraction,  and  then  proceeding  by  this  rule. 

Extract  the  Square  Root  of  fgf. 

S  o  L.— Reduce  gg°  to  f ;  multiplying  both  terms  by  2,  gives  {§,  the 
•square  root  of  which  is  f  nearly. 

4.  <s/272T33    .    .    .  =  164. 

**•   *•  TiI8    •     •  =    I  nearly. 


1.  v  7    .     .     .  =    7  nearly. 

2.  v6f.     .     .=  2}  nearly. 


6.   V90|    .     .  =  94  nearly. 


ART.  392.  Since  the  square  of  a  decimal  has  just  twice  as 
many  decimal  places,  as  the  root  (Art.  384),  the  square  root  of  a 
decimal  must  have  exactly  half  as  many  decimal  places,  as  the 
number  itself;  and  as  the  mode  of  operation  is  the  same  as  in 
whole  numbers,  it  follows  that, 

To  aet  the  square  root  of  a  decimal,  annex  a  cipher,  to  make  the 
number  of  its  decimal  places  even,  (if  it  be  not  so  already) ;  then 
proceed  as  icith  a  whole  number,  pointing  off  from  the  root  half  as 
man;/  dtcimal  places,  as  are  in  the  given  number. 

NOTES. — 1.  The  number  of  decimal  places  must  be  even  before  com- 
mencing the  operation;  otherwise  it  will  be  impossible  to  point  off  in 
the  root  exactly  half  as  many  decimal  places  as  the  number  contains 

2.  The  last,  rule  applies  also  to  mixed  decimals ;  also  to  common 
fractions  or  mixed  numbers,  after  changing  them  to  decimals. 

3.  Since  decimal  ciphers  may  be  annexed  to  any  whole  number  or 
decimal,  if  the  s-juare  root  is  not  exact,  the  process  may  be  continued 
by  bringing  down  two  decimal   ciphers  at  every  step,  and  pointing 
off  one  decimal  place  in  the  root  for  every  pair  of  ciphers  annexed; 
and  the  further  the  process  is  carried,  the  nearer  the  result  is  to 
the  exact  square  root.     The  limit  of  the  emr  is  always  one  of  the 
lowest  order  !»  the  root. 


RAY'S    HIGHER    ARITHMETIC. 


1.  Extract  the  square  root  of  .07625.  Ans.  .276  + 

2.  Of  g  in  decimal  hundredths.  Ans.   .93  + 

3.  Of  2.135  to  thousandths.  Ans.  1.461  + 

4.  Of  3  to  six  decimal  places.  Ans.  1.732051  — 

OPERATION. 


CONTRACTED    METHOD. 


3|  1.732051  nearly. 


1 

1 

27 

200 
189 

27 

200 
189 

343 
3462 
34640 

1100 
1029 

343 

3*4 

1100 
10-29 

I 

! 

5 

7100 
6924 

% 

7100 
(5921 

1760000 
1732025 

176 
173 

346410  I    27975 

RULE    FOR    CONTRACTED    METHOD. 

Extract  the  square  root  as  usual,  until  one  more  than  half  oj 
the  figures  required  in  the  root  have  been  determined;  to  obtain 
the  remaining  figures,  divide  the  last  remainder  by  the  last  divisor, 
using  the  contracted  method  in  Art.  158. 


5. 
6. 
7. 
8. 
9. 
10. 
11. 
12. 
13. 
14. 

V-0081  .  . 

.    .     =.09 
.     .  =.672 
.    =  .298+ 
.    .  =4.21 

nQfH»_l_ 

15. 
16. 
17. 
18. 
19. 
20. 
21. 

22. 

7^    . 

734f 

.    .    =  .92582+ 
.    .    =  5.8843+ 
.    .    =.94868+ 
.    .    =2.5298+ 

7-451584  . 

7-08894  . 
7177724T  . 
7  139:31655 

7!  •  •  = 

i^: 

.8165  nearly. 
.    .  =.175 
=  3.60555+ 
=  1.7724+ 
.     =  .0025 

71089 

».     .    .    =35937 
.    .    =181.02- 

7-030625  . 
713  .  .  . 

76102815944  = 
78120.52+ 

7504125310742198  = 
22452735+ 

73.1415926 
7.^0000625 

APPLICATIONS  OF  SQUARE  ROOT. 
ART.  393     To  find  the  side  of  a  square  figure  of  Riven  area. 

RULE.  —  Reduce  th*  area,  if  necessary,  to  nquare  units;  the 
square  root  of  the  number  thus  obtained  will  be  the  side  of  the 
square  in  linear  units  of  tlie  name  name 


APPLICATIONS  OF   SQUARE   ROOT.  3JP? 


What  is  the  side  of  a  square  field  of  10  acres? 

SOL.— 10  A.  =  1000  sq.  rd.;  then  ./ItiuO  sq.  rd.  =  40  rd.  Ant. 

1.  Find  the  side  of  a  square  field  of  I'J'J  A.  ii  R.     Aits.  140rcL 

2.  The  fencing  fur  a  square  field  of  8  A.  *2  11.  J)  i>.  Ans.  148  rd. 

3.  A  has  5  A.  1 11.  7  1>. ;  7  A.  '211. ;  and  2  A.  3  11.  131*.  of  land, 
ill  a  square:  how  much  fencing  will  enclose  them?  Aim.  200  rd. 

ART.  394.  A  triangle  is  the  space  bounded  by  three  stiaigh 
lines,  called  its  nitlvs. 

A  riyhl-anyleil  triangle  has  two  sides  forming  a  right  angle; 
that  is,  perpendicular  to  each  other. 

The  longest  side  of  a  right-angled  triangle  is  called  the  hypote- 
nn.ie;  the  other  two,  the  perpendicular  sides. 

ART.  395.  To  find  the  hypotenuse,  when  the  perpendicular 
sides  are  known,  use  this 

lln.K. — Square  each  of  the  given  sides;  add  the  results,  and 
extrude  llie  square  root  of  the  sum. 

If  the  perpendicular  sides  of  a  right-angled  triangle  are  3  in. 
and  4  in.,  what  is  the  hypotenuse? 

SOL. — The  square  of  3  is  9;  the  square  of  4  is  16:  their  sum  ia 
25,  the  square  root  of  which  is  6  in.,  the  hypotenuse. 

ART.  396.  To  find  cither  of  the  perpendicular  sides,  when 
the  other,  and  tho  hypotenuse,  are  known,  use  this 

Re* I.E. — Square  each  of  the  given  sides;  take  the  difference  oj 
the  result*,  and  extract  Us  square  root. 

If  the  hypotenuse  is  13,  and  one  of  the  perpendicular  sides  5, 
what  is  the  other  side  ? 

SOL. — The  square  of  13  is  169;  the  square  of  5  is  25:  the  differ- 
ence is  144,  whose  square  root  is  12,  the  other  side. 

1.  Find  the  length  of  a  ladder  reaching  1*2  ft.  into  the  street, 
from  a  window  30  ft.  high.  Aits.  3*2.31  -f-ft. 

2.  What  is  the  iliaijonul,  or  line  joining  the  opposite  corners,  of 
a  square  whose  side  is  10ft.?  Aux.  14  i4xifft. 

?».  What  is  saved  l>y  following  tho  diagonal  instead  of  the  sides 
(01)  and  JKi  rd. )  of  a  rectangle  ?  Ans.  4(5  rd. 

4.  A  boat  in  crossing  a  river  500yd.  wide,  drifted  with  the  cur 
rent  3UOyd. ;  how  far  did  it  go?  AILS.  GIG  f  yd. 

Al<T.  397.  It  is  known,  that  similar  figures  are  to  each  other 
M  the  squares  of  their  like  dimensions  ;  hence, 

1st.  The  ratio  of  the  area*  of  tiro  similar  Jiyiires,  in  equal  tc 
the  square  of  Hie  ratio  of  ani)  tta>  like  dimetiSHitt*  of  lln-in. 

2d.  The  ralit>  of  an;/  two  like  dimension*  of  hco  similat 
figures,  is  equal  In  llie  square  root  of  (he  ratio  o)'  llieir  areas. 


RAY'S    HIGHER   ARITHMETIC. 


One  square  has  a  side  Xi,  times  as  large  as  another:  how  many 
times  doe°  it  eouta:D  >lie  smaller? 

S  OL. — S'ncfc  3*  or  '5'  's  the  ratio  of  the  sidrs,  Vg9,  is  the  ratio  o( 
their  mvas,  sad  the  lar-jer  contains  the  smaller  ^j"  =  1 1 .1 1  times. 

1.  Ont>  square  is  r2j  times  another:  how  many  times  loos  the 
side  of  the  1st  contain  the  -side  of  the  2d?  Jus.  3\. 

"2     The  diagonals  of  two  similar  rectangles  are  as  5  to  12  :  how 
jany  times  does  the  larger  contain  the  smaller?          Aim.  5^  5. 

CUBE  KOOT. 

Am.  398.  The  cube  root  of  a  number  is  another  number, 
which,  being  cubed,  will  produce  the  given  number. 

The  cube  root  of  any  number  containing  three  figures  or  less,  ia 
found  by  trial  and  proof;  thus,  ^/512  is  8,  because  83  =  512. 

If  the  number  does  not  have  an  exact  cube  root,  as  2-17,  it  is  called 
an  imperfect  cube;  and  its  cube-root,  is  obtained  by  trial,  to  within 
unit}/-,  thus,  as  247  lies  between  216  and  343,  iw  cube  root  must  lie 
between  thuirs,  6  and  7. 

ART.  Ji99.  If  a  number  has  more  than  thr?e  figures,  its  cube 
root  will  hnve  two  figures,  tens  and  units;  in  order  to  extract  its 
cube  root,  first  prove  this 

PROPOSITION.  —  Tlie  cube  of  any  number  containing  tens  and 
units,  will  consist  of  the  cube  of  the  tens,  three  times  the  square 
of  the  /en*  multiplied  by  the  units,  three  times  the  tens  multiplied 
by  the  square  of  the  units,  and  the  cube  of  the  units. 

DEJI. -Multiply  the          (47)2  =  1GOO  +     560  +    49 
square  of  47  (Art.  389)  4-'     _  40  -I         7 

by  47  =  40 +7.     The  ~- -— t 1 

result  is  the  cube  of  47,  11200  +  3U20  +  343 

which  consists  of  64000,     64000  +  22400  +  19 0 0 

(the  rube  oj  the  ten*  40),      G4000  +  33000  +  5880  +  343 
of  33000,    (3   times    the 

square  of  the  tens  multiplied  by  the  units  3  X  1^OO  X  7),  of  5^80,  (3 
tinifs  the  tens  multiplied  by  the  square  of  the  v>i'a  3  X  40  Y  4^\  and 
3-13.  (thf  cube  oj  the  units j :  and  the  same  may  be  «J»nru  of  aay  nun>' 
ber  containing  tens  and  units. 

Find  the  Cube  Root  of  238328.  23«-3«»»(IJ2 

A  5  A  L. — Si  n  ce  238328  210* 

ha,  more, hun  3  figures,        80x300=10800 
its  cube  root  must  con-                              _ 
Bist  of  tens  nnd  units; 
and  238328  must  con-  -i  X  2  — 4_ 


lain  the  4  parts  men-  1  1  164 


2232P 


CUBE  ROOT.  339 


tionod  in  the  last  proposition,  the  first  of  which  is  th->  cuf*  of  the 
mi*  Jiyvff  of  the  root.  The  cube  of  tens  being  thousand*,  point  off  tho 
three  right  hand  figures,  and  regard  only  the  238,  which  are  tliou 
sands.  Since  this  lies  between  210  and  343,  its  cube  root  n.ust  lie 
between  theirs,  being  greater  than  the  former  (G)  and  le.-s  than  tho 
latter  (7);  hence,  G  is  the  tens'  figure  of  the  root,  7  being  too  large. 

Cubing  the  tens'  figure,  G  (tens),  gives  21G  (thousands);  subtract 
(his  from  the  given  number;  the  remainder,  22328,  must  be  the  other 
f,  parts  mentioned  in  the  proposition ;  as  the  3d  and  4th  parts  are 
small  compared  with  the  2d,  neglect  them  for  the  present,  and  con- 
sider 22328  as  the  2d  part  alone;  that  is,  3  times  the  square  of  Ike  tens 
multiplied  by  the  units. 

Divide  22328  by  3  times  the  square  of  the  tens  (3  X  3600),  to  get 
the  units'  figure  2,  which  must  not  only  be  placed  in  the  root,  but 
also  used  to  complete  the  divisor,  so  that  when  it  is  multiplied  by  the 
units'  figure  of  the  quotient,  the  product  shall  contain  the  3d  and 
4th  parts  mentioned  in  the  proposition,  as  well  as  the  2d  part. 

To  do  this  conveniently,  increase  the  trial  divisor  (10800)  by  3 
limes  the  tens  by  the  units  (6  X  2  X  30)  and  also  by  the  square  of  the 
units'  (2  X2);  for  these,  when  multiplied  by  the  units'  figure  (2), 
produce  the  3d  and  4th  parts  that  must  be  taken  account  of. 

ART.  400.  The  same  process  and  analysis  apply  to  any 
number;  hence, 

TO  EXTRACT  THE  CUBE  ROOT  OF  A  WHOLE  NUMBER, 

Ilrt.E. — 1.  Separate  the  number  into  periods  of  3  figures,  com- 
menting at  units;  the  left  hand  period  may  contain  1,  2,  or  3 
Jit/tires. 

2.  Find  the  cube  root  of  the  nearest  perfect  cube  below  the  lefl 
hand  period;  thin  will  be  the  \stfignre  of  the  root. 

3.  Cube  the  \stfigure  of  the  root,  subtract  it  from  the  left  hand 
period,  and  briny  down  the  next  period;  this  will  be  the  Jirst 
dividend. 

4.  Take  3  times  the  square  of  the  \stfigure  of  the  root,  annexing 
2  ciphers.  f»r  a  trial  dicisor ;  see  haw  vjten  it  is  contained  in  the 
fiist  dicidend ;  the  quotient  will  be  the  "id  figure  of  the  root. 

5.  Under  the  trial  divisor,  write  3  times  the  2dfi</ure  of  the  rooi 
multiplied  by  the  Jif/nre  before  it,  annexing  one  cipher,  and  nine 
t/ie  t«piare  of  the  second  figure  of  the  root,  and  add  these  it  t/.4 
ti  ial  dicisor  for  a  complete  divisor. 

6.  Mnllififi/  the  complete  divisor  by  the  2<Z  figure  of  the  root, 
and  subtract  the  product  from  (he  first  dicidend,  briny  ing  dowr, 
the  next  period;  this  will  be  the  2</  dividend. 

7.  Take  3  times  the  square  of  that  part  of  (he  root  already  found 
for  a  trial  divisor,  find  another  figure  of  the  root,  complete  the 
divisor,  and  so  on,  until  all  the  periods  have  been  brought  down : 


310 


RAY'S   HIGHER   ARITHMETIC. 


if  Ihf.re  is  no  remainder,  the  exact  cube  root  is  obtained,  but  if 
tlicre  is,  it  is  the  cube  ruul  to  within  unity. 

NOTKS. —  1.   If  any  product  is  greater  than  the  dividend  above  it, 
the  last  figure  of  the  root  is  loo  large. 

2.  If  any  trial  divisor  is  not  contained  in  its  dividend,  put  n  cipher 
(ti  the  root,  two  ciphers  at  the  righ1.  of  the  trial  divisor,  bring  down 
another  period,  and  proceed  as  before. 

3.  If  any  remainder  is  larger  than  the  previous  divisor,  it  does  not 
follow  th.it  the  last  quotient  figure  is  too  small,  unless  the  remainder  is 
large  enough  to  contain  3  times  the  square  of  that  part  of  the  root  already 
found,  with  3   times  that  part  of  the  root,  and  1  more;  for  this  is  the 
proper  divisor  to  go  into  the  remainder,  if  the  root  is  increased  by  1. 

ANS. 

1.  V5T2"     .     .     .     .  =8. 

2.  V 10083"      .     .      =27. 

3.  V  7301384  .     .    =  194. 

4.  V048188I6      •    =450. 


5.  V 1067462048.  =  1022. 


0.  ^6506321  •    =  187— 

7.  VOS705-    .  =  2137— 

8.  */7825    .     =00370+ 

9.  V (10004409373)]  =  13 
10.  V(30S4097945G)i=56 


TO    FIND   THE   CUBE   ROOT   OF   A   COMMON   FRACTION, 

ART.  401.  RIM.E. — It  educe  the  fraction  to  Us  lowest  terms; 
make  the  denominator  a  perfect  cube,  if  it  be  not  so,  by  nuit/i- 
Ph/'n'.l  l'"th  terms  by  the  square  of  the  denominator  or  some  smaller 
number  that  will  answer  the  purpose ;  extract  the  cube  root  of  the 
numerator  for  a  numerator,  and  the  cube  root  of  the  denominator 
for  a  denominator. 

AMS. 

•        '        il 


1. 

2. 

3. 


1+ 


ART.  402.  The  cube  of  ft  decimal  has  just  Iliree  limes  as 
an}'  deetuuU  places  as  the  root;  hence,  the  cube  root  of  a  deci- 
n»al  must  contain  just  one-third  as  many  decimal  places  as  the 
nuu. be r  itself;  hence, 

TO   FIND   THE   CUBE   ROOT   OF   A    DECIMAL, 
Rri.E. — Make  the  number  of  decimal  places  divisible  by  3,  if 
they  be  not  so,  by  annexing  ciphers;  then  proceed  as  in  whole 
numbers,  and  point  oj)'  from  the  root,  one-third  as  many  decimal 
i-lacex  as  are  in  ifie  given  number. 


CUBE  ROOT. 


341 


NOTES. — 1.  The  number  of  decimal  places  roust  be  exactly 
divisible  by  3,  otherwise,  it  would  be  impossible  to  point  off  iu  the 
root,  exactly  3  as  many  decimal  places  as  are  in  the  number. 

2.  This  rule  applies  also  to  mixed  decimals,  aud  lo  common  frnc- 
tions  after  changing  them  to  a  decimal  form;  to  extend  the  process,  if 
the  cube  root  is  not  exact,  bring  down  3  decimal  ciphers  at  every  step 
Bud  make  a  decimal  place  in  the  root,  for  every  3  ciphers  annexed. 


What  is  3/18!>.:*7f>4  to  4  dc 
1S0.375400(5.742G 
1125 

cimal  place? 

CO-NT  R 

1 

1 

? 

ACTED    METHOD. 

^0.375400(5.7426 
25 

75006 
1050 
400 

1375 
3193 

7500 
1050 
4'J 

8599 

974707 
6S4< 
1C 

04375 
00193 

8590 

4182400 
',920224 

97-1700 
GS40 

it; 

4182400 

i 
I 
>  3020224 

981556 

•250  176000 

988-42NH 
3444( 
4 

\ 
} 
197754488 

9*Jj$$|  250170 
19031 

698jf 

589 

98S7  7'24-J 

58421512000 

0 
0 
0 

98911002(1 
10335C 
3 

9892202700  5935321  C77G 

ART.  403.    RULE  FOR  CONTRACTED  METHOD. 

Eftrart  the  cube  root  as  usual,  until  one  more  than  haJf  of  the 
ftj'ii'Kn  required  in  (he  n>ol  have  been  ascertained,  and  t/ien,  trith 
Iht  lust  divisor  and  the  la*t  remainder,  proceed  as  in  contracted 
diciaion  to  determine  the  other  Jiyitres  of  the  root,  except  that  two 
fujnres  are  dropped  instead  of  one  from  the  divisor  at  every  step 
and  one  from  every  remainder. 


AN8. 

ANS. 

1 

V5-M8448 

.    .    =  1.72 

7. 

^7*M> 

.     .     =  .315985 

2. 

5/-2218H.041 

.    .     =  28.1 

8. 

Vio- 

.     .  =  2.924018 

3. 

^:T2~o3  .    . 

=  3.1901544- 

9. 

2/lT 

.    .     =  2.22308 

4. 

iy.0079   .     . 

=  .19916324- 

10. 

3/3 

.     .     .  —  87358 

6. 

U  3.0092      . 

=  1.443724 

11. 

.    .      =.64366 

6. 

V2315.G8     . 

.   =  13.234- 

12. 

^38* 

i  .         =  3.38277 

342  RATS   HIGHER   ARITHMETIC. 


APPLICATIONS   OF  CUBK   ROOT. 

ART.  404.  To  find  the  side  of  a  cube  containing  a  given 
solidity,  U.M-  the  following 

Hn.E  — /fed  are  Ike  solidity,  if  necessary,  in  a  denomination  of 
cn/iic  unit*  ;  extract  ihe  cube  root  of  the  rexult,  and  it  trill  lie  t/>6 
side  >>f'  the,  culm  in  linear  units  of  the  same  name  as  Ike  cubic  units 
in  which  the  solidity  is  expressed. 

Find  the  side  of  a  cubical  box  containing  25  bu. 
25bu.  =  25  X  2150.42  cu.  in. =53700.5  cu.  in.; 
V53760\5  =  37.75  in.  =  3  ft.  1]  in. 

1.  Of  a  cube  containing  70  cu.  ft.,  1323  cu.  in.         Ans.  4  ft.  3  in. 

2.  Of  a  cube  equal  to  a  rectangular  solid  14  ft.  5  in.  long,  0  ft.  8  in. 
wide,  3  ft.  -2  in.  high.  Ans.  6  ft,  8.7  in. 

3.  Of  a  cubical  tank,  to  hold  150 hbl.  Ans.  8ft.  7in. 

4.  Find  the  B«I.  ft.  in  1  face  of  a  cube  containing  39304  cu.  ft. 

An*.  115G. 

6.  The  sq.  in.  in  all  the  faces  of  a  cube  containing  83(55427  cu.  in. 

Ans.  247254. 

ART.  405.  Any  two  similar  solids  arc  to  each  other  as  the 
cubes  of  their  like  dimensions  ;  hence, 

1st  The  ratio  of  two  similar  solids  is  equal  to  the  cube  of  Ihe 
ratio  of  any  two  like  iliiiienxtoini. 

2d.  The  ratio  of  any  two  like  dimensions  of  similar  solids  is 
etjnal  to  the  cube  root  of  (lie  ratio  of  Ike  solids. 

1  The  lengths  of  two  similar  solids  are  4  in.  and  50in.:  the  1st 
contains  l»i  on.  in.:  what  does  the  2d  contain?  Ans.  3125Ucu.  in. 

2.  The  solidities  of  two  halls  are  IRHcu.in.  and  875cu.  in.;  the  diam- 
eter of  the  2d  is  11 2  iu.;  find  the  diameter  of  the  1st.  Ans.  10^  iu. 

EXTRACTION  OF  ANY  ROOT. 

ART.  406.  There  is  a  general  method  of  extracting  roots, 
called  llurni-r's  method,  from  its  inventor,  which  has  great  ad- 
vantages. It  is  comprised  in  this 

RTLE    FOR    EXTRACTING   ANY   ROOT. 

1.  Make  as  many  columns  as  there  are  units  in  the  index  of  the 
root  to  be  extracted;  place  (fie  airen  iininlier  at  the  head  of  the  riyht 
hand  column,  and  ciphers  at  the  head  of  the  others. 

2.  Commencing  at  the  riaht,  separate  the  gicen   mnnfar   into 
periods  of  ax  many  faitres  an  there  are  cn/minin  :  extract  the  re~ 
quired  root   to  irif/iiit  unity,   of  the  left-hand  period,  for  the   1st 
figure  of  the  root. 

3.  Add  tlris  (inure  into  the  Ast  coJninn.  mnltiph/  it  tltcn  In/  »/.«•//", 
and  set  it  in  the  "2d  column;    multiply  this  aaain  b;i   the  same 
fujurc,  and  set  it  in  the  '3d  column,  and  so  on,  placing  ihe  last  pro- 


EXTRACTION  OF  ANY  ROOT. 


343 


duct  in  the  right  hand  column,  under  Ihnf  part  nf  (he  gircn  in/in- 
ber  from  which  the  Jigure  was  derived,  and  subtracting  it  J'roin  tlu 
Jignres  above  it. 

4.  A<l<l  the  same  figure  in  In  the  \st  column  again,  multiply  the 
"esnll  by  the  fgnre  again,  adding  the  product  to  the  2d  coining 
and  so  ou,  stopping  at  the  next  to  the  last  column. 

5.  Jfepeat  this  process,  leaping  off  one  column  at  the  right  every 
time,  until  all  the  columns  have  been  thus  dropped  ;  then  annex  one. 
tiplter  to  the  number  in  (he  1st  column,  two  to  the  number  in  the  Id 
column,  awl  so  on,  to  the  number  in  the  last  column,  t<>  which  the 
next  period  of  figures  from  the  given  number  must  be  brought  down, 

6.  Picide  (he  number  in  (he  last  column  by  the  number  in  the 
precious  column  as  a  trial  divisor  (making  allowance  for  complet- 
ing the  divisor) ;  this  will  give  the  2<f  .figure  of  the  root,  whii-h 
must  be  nsed  precisely  as  the  l.v/  Jigure  of  the  root  has  been:  and 
so  on  till  all  the  periods  have  been  brought  down. 

ART.  407.     The  process  may  often  be  shortened  by  this 

RCLE  FOR  CONTRACTED  METHOD. —  Obtain  one  lens  than  half  of 
the  Jignres  required  in  the  rout  as  the  rule  directs ;  then,  instead  of 
annexing  ciphers  and  bringing  down  a  period  to  the  last  numbers 
in  /he  columns,  leave  the  remainder  in  the  right  hand  column  for  a 
dividend ;  cut  off  the  right  hand  figure  J'rom  the.  last  number  of 
the  previous  column,  two  right  hand  fgnres  from  the  (oxt  number 
in  tlie  column  before  that,  and  so  on,  always  culling  ojf'  one  more 
Jiyure  fur  every  column  to  the  left. 

With  the  number  in  the  right  hand  column  and  the  one  in  the 
precious  column,  determine  the  next  Jiyure  of  the  root,  and  use  it 
as  directed  in  the  ni/e,  recollecting  that  the  fgures  cut  ojf  are  not 
used  except  in  carrying  the  tens  they  produce.  This  process  is  con- 
tinued until  the  required  number  of  figures  are  obtained,  observ- 
ing that  irhi.n  all  the  Jignres  in  the  but  number  of  any  column  are 
cut  off,  that  column  will  be  no  longer  used. 

REM. — A»l«l  to  the  1st  column  mentally;  multiply  and  n<l«l  (o  the 
next  column  in  one  operation  :  multiply  aiiJ  subtract  from  the  right 
band  column  iu  like  man  HIT. 

Extract  the  cube  root  of  44.6  to  six  decimals. 

A  i  «nn/Q    t  1TQOQ         EXP.— The  trial 

17600  iivi8or8 

1725000 
238136 
12182 
805 
111 


0 
3 
6 

90 

95 

100 


0 
9 

2700 
3175 
367500 
371716 
1050  375D43 
1054  37659 
1058  3772£ 


known  by  ending 
in  two  ciphers;  the 
complete  divisors 
stand  just  beneath 
them.  After  get- 
ting 8  figures  of 
tLe  root.  «  on  tract 
the  operation  by 
last  rule. 


044 


RAY'S   HIGHER   ARITHMETIC. 


Extract 

the  5th 

root  of  1250 

,  to  4  places 

of  decimals. 

0 

0 

0 

0 

1256  (  4.1668 

4 

16 

64 

256 

23200000 

8 

48 

256 

12800000 

97-13799 

12 

96 

640000 

13456201 

1014765 

16 

16000 

656201 

1412880^ 

113733 

200 

16201 

672604 

1454839 

201 

16403 

689210 

1497403 

202 

16606 

6993 

150172 

203 

JlWP 

7094 

150604 

20-1 

nn 

RE»I.— We  have  seen  (Art.  382)  that  (43)5  =4T6,  that  is,  if  4  b« 
raised  10  the  3d  power,  and  this  product  raised  to  the  5th  power,  the 
result  will  be  the  loth  power  of  4 ;  hence,  conversely  if  the  5th  -oot 
of  4  '  °  tie  taken,  ami  (lie  3d  root  of  this  result,  it  will  give  4.  which 
is  the  loth  root  of  4 '  °,  that  is,  the  3d  root  of  the  5th  root  of  a  num- 
ber is  equal  to  the  15th  root  of  that  number.  Hence, 

RCI,E. —  Whenever  the  index  of  the  mot  to  be  extracted  ran  be 
separated  into  factors,  extract  success! rely  the  roots  indicated  by 
those  factors,  and  Ike  final  result  thus  obtained  will  be  Ike  root 
required. 

In  using  this  rule,  begin  with  the  smallest  roots. 
Thus,  to  get  the  4th  root,  extract  the  square  root  of  the  square  root. 
And  to  get  the  Oth  root,  extract  the  cube  root  of  the  square  root. 
Ami  to  get  the  8th  root,  extract  the  sq.  root  3  times  in  succession. 
Arid  to  get  the  Oth  root,  extract  the  cube   root  of  the  cube  root, 
and  so  on. 


1.  V'JT.41    . 

2.  V3_._I 

3.  5/21035.8 


MM. 

.    =3.1416 

4.     $/  .0004850 

.    =  1.2457 

5.  vi;  .  . 

.      =  5.254 

o.    V5f.  . 

ASS. 

=  .0786007 

,    =  .83938 

=  1.87445 


XXXV.    SERIES. 

AUT.  408.  A  SERFES  is  a  succession  of  numbers,  each  derived 
from  the  priM-ediii<i.  arvordinj;  to  a  fixed  law. 

The  iiumliors  which  form  a  scries  are  railed  its  terms;  if  they 
increase  toward  the  rijrlu,  it  is  an  Asceuilimj  series  :  if  they  de- 
crease toward  the  ri<;ht,  it  is  a  Descending  series:  the  first  and  last 
terms  are  the  extremes.  Series  are  Arithmetical  or  Geometric. 


ARITHMETICAL  SERIES.  345 


ARITHMETICAL  SERIES. 

ART.  409.  An  Arithmetical  series  is  one  whose  terms  increase 
or  deureii.-L'  l»v  the  addition  or  rabtraetian  of  a  fixed  nuiiilii-r,  ralhnj 
the  comiiKin  difference ;  as,  'Z,  5,  8,  &c.;  or,  10,  7,  4  ;  iu  which 
the  common  difference  is  3. 

ART.  410.  To  find  the  last  (or  any)  term  of  an  arithmetical 
series,  when  the  1st  term,  common  difference,  and  number  of  tcrma, 
are  known. 

RITI.E. — Multiply  the  common  difference  by  1  less  than  tin  num- 
ber of  terms,  and  add  the  product  to,  or  subtract  it  from,  the  1*1 
term,  according  as  the  series  is  ascending  or  descending. 

1.  Find  the  12th  term  of  the  series  3,  7,  11,  &c.  Ann.  47. 

2.  The  18th  terra  of  the  series  100,  9G,  &c.  Ant.  32. 

3.  The  64th  term  of  the  series  3.j,  6f,  &c.  An*.  11-  $ 

4.  The  10th  term  of  the  series  .025,  .037,  &c.  Ans.  .133 

ART.  411.  To  find  the  1st  term,  when  the  last  term,  common 
difference,  and  number  of  terms,  are  known. 

II r I.E. —  Transpose  the  series,  and  then  find  its  last  term;  this 
trill  be  the  1st  term  of  Ike  given  series. 

JCojE. — If  the  given  series  is  ascending,  the  transposed  one  will 
be  descending,  and  vice  versa. 

1.  Find  the  1st  term  of 68,  71,  having  19  terms.          Ans.  17. 

2.  Of  ....117,  123!,,  130,  having  6  terms.  Ann.  97  £ 

3.  Of  ....18|,  12A~  G],  having  365  terms.  Ant.  2281 } 

ART.  412.  To  find  the  common  difference,  when  the  extremes 
and  number  of  terms  are  known. 

RU.B. —  Divide  the  dijferenct  of  the  extremes  by  the  number  of 
term*  less  one. 

1.  Find  the  common  difference  of  a  series  whose  extremes  nrc  8 
ami  28,  nnu  number  of  terms,  6.  Ans.  4. 

2.  Extremes  are  4i  and  20|,  and  number  of  terms,  14.    Ans.  1 } 

3.  44th  term  is  150,  and  19th  term,  30.  Am.  4i 

4.  4th  term  is  7,  and  14th  term,  39.  Ans.  b£ 

ART.  413.  To  find  the  number  of  terms,  when  f.he  extremes 
tnd  common  difference  are  known. 

Rri.E. — Dieide  the  difference  of  the  extremes  by  the  common 
difference,  and  add  1  to  the  quotient. 

1.  What  is  the  number  of  terms  in  a  series  whose  extremes  nre  9 
tud  42,  ami  common  difference,  3?  Anx.  12. 

2.  Whose  extremes  are  3  and  10A,  and  com.  dif.  |?        Ans.  21. 
3    In  the  series  10,  16  ...  600 1~  AM.  99. 


340  RAY'S  HIGHER  ARITHMETIC. 


A  RT.  414.  To  insert  a  given  number  of  arithmetical  meant 
between  two  given  numbers. 

ilci.E. —  Take  the  yicen  numbers  as  the  extremes,  and  2  more  than 
Hie  number  of  meant  as  the,  number  of  term*,  "fan  arillnnetiial 
scries,  and  Jind  the  common  difference  by  rule  in  Art.  412.  Add 
tit  is  ••nmi/tun  difference  to  the,  smaller  number,  to  gel  the  1st  mean; 
add  it  to  the  \st  mean,  to  yet  the  °2d  mean,  and  so  on. 

1.  Insert  1  arithmetical  meim  between  8  and  54.  Ans.  31. 

2.  5  arith.  means  between  G  and  30.          Ans.  10,  14,  18,  22,  26. 

3.  2  arith.  means  between  4  and  40.  Am.  16,  28. 

4.  4  arith.  means  between  2  and  3.  Ans.  2g,  2|,  2»,  2|. 

ART.  415.  To  find  the  sum  of  the  terms  in  an  arithmetical 
series,  when  the  extremes  and  number  of  terms  are  known. 

RULE. — Multiply  half  the  sum  of  the  extremes  by  the  number  of 
terms :  the  product  will  be  the  sum  of  the  series. 

1.  Find  the  sum  of  the  arithmetical  series  whose  extremes  are  850 
and  0,  and  number  of  terms,  57.  Ans.  24225. 

2.  Extremes,  100  and  .0001 :  No.  of  terms,  12345.  Ans.  617250.G1725 

NOTE. — It  may  be  necessary  to  find  the  extremes  or  number  of 
terms  by  one  of  the  previous  rules,  before  applying  this  rule. 

3.  What  is  the  sum  of  the  arithmetical  series  1,  2,  3,  &c.,  having 
10000  terms  ?  Ana.  50000000. 

4.  Of  I,  3,  5,  &c.,  having  1000  terms?  Any.  1000000. 

5.  Of  999,  888,  777,  &c.,  having  9  terms?  Ans.  41)95. 

6.  Of  4.12,  17.25,  30.38,  &c.,  having  250  terms  ?    Ans.  409701 .25 

7.  Whose  5ih  term  is  21 ;  20th  term,  CO;  number  of  terms,  46  ? 

Ant.  3178|. 

GEOMETRIC  SERIES. 

ART.  416.  A  GEOMETRIC  SERIES  is  one,  in  which  each  term 
ft  formed  by  multiplying  the  previous  one  by  a  fixed  number 
sailed  the  common  ratio. 

ART.  417.  To  find  the  last  (or  any)  term  of  a  geometric  series, 
when  the  lr>t  term,  the  common  ratio,  and  the  number  of  terms 
arc  known. 

IJt'i.E. —  liaise  the  common  ratio  to  a  power  whose  degree  is  on4 
less  than  the  number  of  terms,  and  multiply  the  I  at  term  by  it. 

NOTE. — The  commro  ratio  in  a  given  geometric  series  may  b« 
found  by  dividing  any  term  by  the  preceding. 

1  Find  the  last  term  in  the  series  <>  1,  32,  &c.,  of  12  terms.    Ant.  3'2 

2.  In  2,  o,  12.J,  &c.,  having  6  terms.  Ans.  195 /g 

8.  In  100,  20,  4,  &c.,  having  9  terms.  ^"•'•15623 
4.  1st  term,  4  ;    com.  ratio,  3:    find  the  10th  term.    Ans.  78732 


GEOMETRIC   SERIES.  347 


5.  3d  term,  1C;  com.  ratio,  6:   find  the  9th  term.      Ana.  74f»4'.)6. 

6.  83d  term,  10lM;  com.  ratio  |:  find  »,he40lh  term.    AM.  13u',£. 

ART.  418.  To  find  the  1st  term,  when  the  last  term,  nuiubei 
of  terms,  and  common  ratio  are  kuown. 

Un.E. —  Transpose  the  series ;  fnd  the  last  term  by  the  previous 
rule  :  thin  trill  be  the  1st  term  of  tlie  given  series. 

NOTE. — The  common  ratio  of  the  transposed  series  will  be  the 
frHiiinoii  ratio  of  the  given  series  inverted. 

1.  Find  the  1st  term  of  the  series  ...90,  180,  of  6  terms.   Ans.  6f . 

2  Of  ...27u"t9o.  f  005'  having  H  terms.  Am.  j,1^ 

3.  9th  term,  570  ;  com.  ratio,  3:  find  the  1st  term.        Ans.  f-.^ 

4  18th  term,  1 5625;  com.  ratio,  %2^:  find  the  10th  term.  Am.  lO./j 

5  6Gth  term,  1410;  com.  ratio,  6:  find  the  49th  term.   Ans.  §73 

ART.  419.  To  find  the  common  ratio,  when  the  extremes  and 
number  of  terms  are  known. 

RUI.B. —  Dicide  (he  laat  term  by  the  frst,  and  extract  that  root 
of  (lie  ifiiofii-nt  u-ltoxe  index  is  one  less  than  the  number  of  terms' 
i/iix  if  ill  be  the  common  ratio. 

1.  Find  the  common  ratio:   1st  term,  8;  4th  term,  612.      Ans.  4. 

2.  1st  term,  4Jg  ;  llth  term,  49375000000.  Ans.  10. 

3.  IGlh  term,  729;  22d  term,  1000000.  AM.  3j. 

ART.  420.  To  insert  a  given  number  of  geometric  means  be- 
tween two  given  numbers. 

RULE. —  Take  the  giren  numbers  as  the  extremes  of  a  geometrical 
series,  and  '£  more  than  the  number  of  means  for  the  number  of 
terms,  and  Jiiid  the  common  ratio  by  the  la*l  rule;  multiply  the 
Jirst  term  by  the  common  ratio,  to  gel  the  \xt  mean;  multiply  this 
by  the  common  ratio  to  get  the  '2d  mean,  and  no  on. 

1.  Insert  1  geometric  mean  between  63  and  112.  Ans.  84. 

2.  4  geometric  means  between  6  and  IW         Ans.  12,  24,  48,  96. 

3.  3  geometric  means  between  jgsin  atl1'  g-  Ans.  55*53,  5 73,  7*2 

4.  2  geometric  means  between  14.08  and  3041.28. 

Ana.  84.48  and  506.88 

ART.  421.  To  find  the  sum  of  all  the  terms  in  a  geometric 
series,  when  the  extremes  and  common  ratio  are  known. 

Rrr.E. — Multiply  the  last  term  by  the  common  ratio;  jind  the 
difference  between  this  product  and  the  firxt  term,  and  divide  it  by 
the  difference  between  the  common  ratio  and  1. 

NOTE  — It  mny  be  necessary  sometimes  to  find  one  of  the  extremei 
«y  one  of  the  previous  rules. 

1.  Find  the  sum  of  6,  12,  24,  &c.,  to  10  terms.  Ans.  0138. 

2.  Of  10384,  8192,  &o.,  to  20  terms  Ans.  82767  $  j 
8.  Of  -jj   |,  287,  &c.,  to  7  terms.  Ans.  1  j'f  |^ 


348  RAY'S   HIGHER   ARITHMETIC. 

4.    Extremes,  25  nn.l  102100;     No.  of  levins,  13.        Sum,  204 
6.    Int  term,  7;    IJih  term,  1701  :     No.  of  terms,  9.      Sum. 

6.  3-1  term,  40;    List  ( I  lib)  term,  2r,21440.  Sum,  3J!f>2VJ\. 

7.  4th  term,  216;  8th  term,  4'2'j  ;  No.  of  terms,  10.  Sum,  214(J2'7. 
In  an  infinite  decreasing  geometrical  series  the  last  term  is  0. 
Find  the  puai  of  the  following  infinite  geometrical  series : 

8.  Of  1,  £,  4,  &c.     Am.  2.   10.  Of  £,  |.  ^  &c.   Ant.  2. 

9.  Of  f,  v/5,  VVk,  &c.  ^n«.  1  A.   11.  Of  I,  1,  !),  &c.    Ans.  8j. 

12.  Of  .36  =  .3636,  &c.  =  -j30fi(j  +  7  oVo  fi»  &c-        -A  *».  ij- 

13.  Of  .349206,  of  480,  of  6.         AM.  § 5  and  ]f  J  and  | 


XXXVI.    PERMUTATIONS. 

ART.  422.  PERMUTATIONS  arc  the  changes  of  order  which  a 
nun) her  of  things  may  undergo. 

To  find  the  number  of  permutations  possihle  with  a  given  num- 
ber of  objects,  using  them  all  each  time. 

Rri.E — Multiply  tot/fiber  the  series  of  natural  numbers  fwm 
1  to  the  yi  ceii  number  ofolyects  inclusive. 

1.  How  long  can  7  persons  sit  in  different  order  at  lable.  allowing 
365]  da.  to  a  ve:ir,  nnd  3  meals  a  day?  Ana.  Ayr.  l'l!'dn. 

2.  How  in  any  rhanjres  of  order  are  possible  with  the  letters  that 
compose  the  word  antlie-mf  Ana.  720. 

3.  How  ninny  different  numbers  can  be  expressed  by  the  same  fig- 
ures as  the  number  1234"»678'.tO?  A H«.  302X800. 

4.  In  how  many  different  ways  can  the  8  notes  of  nn  octnve  be 
written?  A ns.  40320. 

ART.  423.  To  find  the  numher  of  permutations  of  a  given 
number  oi  objects,  using  a  given  number  less  than  all,  each  time. 

Riri.E. — .Ifitl/ifiti/  toy  el  her  the  series  of  natural  numlierx.  c«m- 
meii'-in'j  irith  1  itmrc  than  the  number  of  objects  omittftl  in  each 
permutation,  and  ending  with  the  ichvle  number  of  objects. 

1.  How  many  permutations  can  be  made  of  7  letters,  using  3  each 
time?  Ans.  210. 

2.  Of  5  letters,  using  2  each  time?  Ans.  20. 

3.  Of  G  letters,  using  1  each  time?  Ant  6. 


XXXVII.    COMBINATIONS. 

ART.  424.    COMIIIVATIONS  are  the  different  sets  containing  th« 
same  number  of  objects,  which  may  be  selected  from  u 
larger  number  of  objects. 


Bit  STEMS  OF  NOTATION.  349 

To  find  the  number  of  combinations  possible  with  a  given  num- 
ber of  objects,  using  <i  given  number  each  time. 

llci.E. —  Find  by  the  last  rnle,  the  number  of  permutations  ;xw- 
gilile  icil/i  lite  whole  number  of  objects,  using  (is  man;/  each  lime  as 
are  In  lie  in  each  combination  ;  then  find  by  the  rule  in  Art.  V-2.,  the 
number  oj'  permutations  possible  with  the  number  of  object 's  in  a 
combination,  using  all  each  lime:  the  farmer  result  dicidcd  by  the 
latter  will  gice  the  number  of  combinations  required. 

1.  How  many  combinations  of  lOletters,  taken  7  iiiaset?  An*.  120. 

2.  Of  9  letters,  taken  4  in  each  set?  AM.  12<i. 
&    Of  8  letters,  taken  3  in  a  set;  also,  5  in  a  set?  Am.  Each  50. 


XXXVIII.    SYSTEMS  OF  NOTATION. 

ART.  425.  The  radix  of  a  system  of  Notation,  is  the  number 
of  units  of  each  order  which  make  one  of  the  next  higher  order. 

The  radix  of  the  ordinary  or  decimal  system  is  10.  Other  systems, 
tunary,  ternary,  &c.,  arise  from  other  radices ;  as  2,  3,  &c. 

ART.  428.  To  change  a  number  in  the  decimal  system,  to  any 
other  system  whose  radix  is  known. 

RCI.E. —  Diride  (fie  given  number  l>y  the  radix  of  tlie  system  to 
trhich  it  in  to  he  reduced ;  divide  thin  quotient  by  (he  radix  again, 
uiid  so  on,  until  a  quotient  is  obtained  smaller  than  Hie  radix : 
the  last  quotient,  icilh  the  sererul  remainders  annexed,  will  be  the 
number  in  the  required  system. 

NOTE. — When  there  is  no  remainder,  place  a  cipher  instead. 

1.  Change  87r>-l  in  the  decimal  system,  to  the  binary  ("2).  quinary  (5), 
and  nonary  (9)  systems.  Ana.  10001000111100;  240024;  13o'l7. 

ART.  427.  To  reduce  a  number  in  any  system  whose  radix  is 
known,  to  the  decimal  system. 

RULE. — Multiply  the  left  hand  figure  of  the  giren  number  by  the 
radix  of  (he  giren  xysfem,  and  to  (he  prod  net  add  the  next  Jin  arc; 
multiply  this  sum  by  (he  radix  again,  and  add  in  the.  next  Jiyurt, 
ami  so  mi,  until  all  the  ^figures  hare  been  added  in :  the  la-st  result 
will  be  the  number  in  the  decimal  system. 

1.  Reduce  70.V53-I1  in  the  octary  (8)  system,  nnd  201221  in  the 
ternary  (3)  system,  to  the  decimal  system.  Am.  1808785 ;  638. 

ART.  428.  To  change  a  number  in  any  system  whose  radix 
is  known,  tu  any  other  system  whose  radix  is  known. 

llri.E. —  AY/-*/  reduce  the  given  number  to  the.  decimal  system  by 
tlie  last  rule;  then,  redact  this  result  to  the  required  system,  by  tkt 
rule  in  Art.  42G. 


350  RAY'S   HIGHER   ARITHMETIC. 

1.  Change  4210532  from  the  senary  (6)  system  lo  the  qunternary 
(4)  system.  Ana.  3orJ3lM20. 

REMARK — Numhers  expressed  in  any  other  than  the  decimal 
system,  may  he  added,  subtracted,  multiplied,  divided,  &c.,  as  in  the 
decimal  system,  except  that  in  currying,  burrowing,  nn<l  reducing,  take 
1  of  <inif  order,  not  equal  to  10  of  the  rtfxt  lower,  but  equal  to  as  many  oj 
the  next  lower,  as  there  are  units  in  the  radix  of  the  system. 


XXXIX.    DUODECIMALS. 

ART.  429.  The  duodecimal  system  is  the  one  whose  radix  id 
12.  It  is  applied  in  practice  to  the  measurement  of  surfaces  and 
solids ;  the  foot,  square  foot,  or  cubic  foot,  is  the  unit ;  the  duode- 
cimal divisions  are  called  primes,  seconds,  thirds,  &c. 

ART.  430.    For  all  operations  in  duodecimals  use  this 

Re  I.E. — Change  the  number  of  whole  feet  from  the  decimal  in  the 
duodecimal  system,  if  necessary,  after  which  net  dutrn  the  duodeci- 
mal divisions,  in  order,  using  Q  J'nr  10,  and  ®  for  I  1,  ami  s< pa- 
rating  the  units  from  the  duodecimal  orders  by  (:)  instead  »f  (.). 

Then  proceed  with  the  operation  as  with,  ordinary  nmnhfrs,  except 
that  carrying,  bin-rowing,  reducing,  tfcc.,  are  performed  on  the  basis 
q/'12  instead  of  10;  the  result  thus  obtained  will  be  »>»  the  duode- 
cimal system,  and  the  number  of  whole  feet  may  then  be  chanyed  to 
tlie  decimal  system  if  necessary . 

REMARK. — In  adding  and  subtracting  duodecimals,  the  number 
of  whole  feet  need  not  be  expressed  in  the  duodecimal  system. 
Primes  are  marked  ('),  seconds  ("),  thirds  ('"),  and  so  on. 

1.  Add  3  ft.  2'  6",  1  ft.  8"  10"',  and  4  ft.  7'  9'".  Ans.  8  ft.  10  3"  7'". 

2.  Subtract  14  ft.  9'  7"  8"'  from  20  ft.  10".          Ann.  5  ft.  3'  2"  4'". 

3.  Multiply  3  ft.  8'  4"  by  7  ft.  2'  4".    Ana.  26  sq.  ft.  6'  11"  5'    4"". 

4.  What  is  the  surface  of  a  board  13  ft.  8'  long  by  1  ft.  9'  3"  wide? 

A  »s.  24  sq.  ft.  ?.'  5". 

5.  Of  a  floor  32  ft.  8'  4"  long  by  21  ft.  6'  8"  wide? 

An.-.  704  sq.  ft.  8'  11"C'"  8"". 

6.  What  is  the  solidity  of  a  log  16  ft.  2'  4"  long  by  1  ft.  9'  wide  and 
10*  6"  thick  ?  Ans.  24  cu.  ft.  9'  0"  10'"  6"". 

7.  Diviue  14  ft.  3*  11"  4'"  by  8.  Ana.  1  ft.  9'  5"  11'". 

8.  Divide  22  ft.  1'  2"  9"'  by  3  ft.  7".  .4ns.  7.2-5  =  7}. 

9.  Divide  53  sq.  ft.  9*  2"  2'"  by  6  ft.  3'  2".  Ant.  8  ft.  7'. 

10.  Divide  99  cu.  ft.  9'  11"  by  17  sq.  ft.  4'  4".  Ant.  5  ft.  9'. 

11.  Divide  04  cu.  ft.  10"  10'"  by  2H  ft.  (>'  2".  Ans.  2  sq.  ft.  5'. 

12.  How  long  must  a  board  be,  that  is  1  ft.  5'  wide,  to  contain  T.t  sq. 
*  1   2'"  7""?  Ann.  13  ft.  5'  7"  IT". 

13.  What  is  the  hight  of  a  rectangular  log  which  contains  1»>5  cu.  fL 
2  2*  8'",  and  is  21  ft.  4'  long  by  3  ft.  T  2"  wide?  Ana.  2  ft.  «'. 

REMAP  K.  —  If  the  pupil   prefers,   he    may  solve  these  examples 
according  to  the  method  explained  in  Ray's  Arithmetic,  Tl*.d 
Art.  276.   similar  to  the  operations  in  compound  numbers. 


MENSURATION.  35] 


XL.    MENSURATION  OF  SURFACES. 

AUT.  431.    A    Parallelogram   is   a  figure   bounded   by   four 
straight  lines,  and  whose  opposite  sides  are  equal  and  parallel. 
If  the  adjacent  Rides  arc  perpendicular  to  each  other,  the  figure  U 
rectangle;  if  they  are  also  equal,  it  is  a  square. 

To  find  tlie  area  of  any  parallelogram,  rectangle,  or  square, 

Rai,E. — Multiply  the  base  and  altitude  together,  after  expressing 
them  in  the  same  denomination :  the  product  will  be  the  area  in 
square  uuita  of  the  same  name. 

NOTE. — The  base  is  any  side;  the  altitude  is  the  perpendicular  or 
shortest  distance  from  the  opposite  side  to  the  base.  In  a  rectangle 
and  square,  two  adjacent  sides  are  the  base  and  altitude. 

1.  Find  the  area  of  a  parallelogram  whose  base  is  0  ft.  4  in.  and 
altitude  2  ft.  5  in.  Am.  22  sq.  ft.  80  sq.  in. 

2.  Of  an  oil  cloth  42  ft.  by  5  ft.  8  in.  AM.  2ii^  sq.  yd. 

3.  How  many  tiles  8  in.  square  in  a  floor  48  ft.  by  10ft.?  Ana.  1080. 

ART.  432.  A  Triangle  is  a  figure  of  '3  sides ;  any  side  is  the 
base ;  the  altitude  is  the  perpendicular  or  shortest  distance  to  the 
base  from  the  opposite  vertex,  or  corner. 

To  find  the  area  of  a  triangle  with  the  base  and  altitude. 

RULE. —  Take  half  the  product  of  the  base  and  altitude,  after  ex~ 
pressing  them  in  the  same  denominations ;  this  will  be  the  area  in 
square  uni/s  of  the  same  name. 

To  find  the  area  of  a  triangle  with  the  three  sides. 

RULE. —  Take  half  the  sum  of  the  sides ;  subtract  each  side  from 
it;  multiply  together  the  three  remainders  and  the  half  sum  ;  ex- 
tract the  square  root  of  the  product ;  this  will  be  the  area  in  square 
units. 

1.  Find  the  area  of  a  triangle  whose  base  is  72  rd.  and  altitude 
16  rd.  Am.  3  A.  2  R.  10  P. 

2.  Base  13  ft.  3  in. ;  altitude  9  ft.  6  in.     Ans.  62  sq.  ft.  135  sq.  in. 

3.  Sides  1  ft.  10  in.;  2  ft. ;  3  ft.  2  in.       Am.  1  sq.  ft.  102— sq.  in. 

4.  Sides  15  rd.;  18  rd.;  25  rd.  Am.  3  R.  13.00— P. 

ART.  433.  A  Trapezoid  is  a  figure  of  4  sides,  two  of  which 
are  parallel  but  not  equal,  and  are  called  the  bases. 

To  find  the  area  of  a  trapezoid. 

RULE. —  Take  half  the  sum  of  the  bases  and  multiply  it  by  the 
altitude,  after  expressing  them  in  the  same  denomination:  the 
product  trill  be  the  area  in  square  units  of  the  same  name. 

NOTE. — The  altitude  is  the  perpendicular  between  the  bases. 

1.  What  is  the  area  of  a  trapexoid  whose  bases  are  9  ft.  and  21  ft 
and  altitude  10  ft.?  Ans.  240  sq.  ft. 

2.  Bases  43  rd.  and  65  rd. ;  altitude  27  rd.?  Ans.  9  A.  18  P. 


352  RAY'S   HIGHER  ARITHMETIC. 

ART.  434.     To  find  the  area  of  any  irregular  figure  bounded 
b     four  or  more  straiht  lines. 


UCI.E.  —  S'-ji-irnfe  the  figure  into  triangles  by  joining  its  angular 
)>niii/:<  ;  find  t.'ne,  area  of  tack  triangle,  and  take  their  sum  fur  l/ic 
a/v«  i.>f  th  e  figure. 

1.  What  is  the  area  of  a  figure  made  up  of  3  triangles  whose  bases 
arc  K>.  r_\  1G  rd.  ami  altitudes  '.»,  10,  10.1  rj.  7  ^ns.  1  A.  1  R.  19  P. 

'2.  Whose  sides  arc  10,  12,  14,  It)  rd.  iri  order,  and  distance  from  the 
Bturliug  point  to  the  opposite  corner,  18  rd.  ?  Am.  1  A.  3.9  —  1'. 


PLASTERERS',  PAINTERS',  AND  PAVERS'  WORK, 

ART.  435.  Is  computed  by  the  sq.  yd.;  glaziers'  work  by 
the  sq.  ft.  or  pane;  carpenters'  and  joiners'  work  by  the  sq.  yd., 
ami  sometimes  by  the  square,  which  is  10ft.  square,  and  con- 
tains lOOsq.  ft. 

1.  Find  the  cost  of  roofing  a  house  GO  ft.  long  and  22  ft.  9  in.  from 
the  ridge  to  the  eaves,  at  3G  ct.  a  sq.  yd.  Anx.  $10'.i.20 

2.  How  much   wainscoting   in  a  room  25ft.  long,  18ft.  wide,  and 
14  ft.  3  in.  high,  allowing  a  door  7  ft.  2  in.  by  3  ft.  4  in.,  and  two  win- 
dows, each  5  II.  8  in.  by  3  ft.  6  in.,  and  a  ch'imney  G  ft.  4  in.  by  5  ft.  6 
in.;  charging  the  door  and  windows  half-work?  Ans.  128^^  sq.  yd. 

3.  How  many  squares  in  a  partition  150  ft.  9  in.  long  by  10  ft.  4  in. 
high?  Ann.  1G.1D75 

4.  Find  the  cost  of  flooring  and  joisting  a  bouse  of  3  floors,  each  48 
ft.  by  27  ft.  deducting  from  each  floor  for  a  stairway  12  ft.  by  8  ft.  3 
in.,  allowing  '.» in.  rests  for  the  joists;  estimating  the  flooring  and  joist- 
ing  h'ticftn  the  walls  at  $1.4G  a  s«j.  yd.,  and  the  joisting  in  the  walla 
at  7G  ct.  a  sq.  yd.;  each  row  of  rests  being  measured  48  ft.  long  by  9 
in.  wide.  Ans.  $(H)0.78 

ft.  What  is  the  cost  of  plastering  a  partition  7  ft.  8  in.  long  and  10 
ft.  3  in.  high,  at  4o  ct.  a  s<{.  yd.,  deducting  a  door  0  ft.  3  in.  bv  2  ft.  10 
in.?  4**.  $3.0-1 

6.  Flow  many  ?q.  yd.  of  plastering  in  a  room  30  ft.long,  2-3  ft.  wide, 
and   12  ft.  high,  deducting  3  windows,  each  8  ft.  2  in.  by  5  ft..  2  doors 
each   7  ft.  by  3  ft.  6  in.,  and  a  fire-plac?  4  ft.  G  in.  by  4' ft.  10  in.;  the 
sides  of  the  windows  being  plastered  15  in.  deep?     And  what  will  it 
cost,  at  25  ct.  a  sq.  yd.?  Aiu.  215.}  s<|.  yd.;  cost  $53.83 

7.  Kind  the  cost  of  painting  a  wall  14  ft.  by  9i  ft.,  at  18  ct.  a  sq 
yd.,  except  a  chimney  4  ft.  0  in.  by  3  fu  10  in.  Ant.  $2.31  .J 

8.  How  much  painting  on  the  sides  of  a  room  20  ft.  long,  1  I  ft.  G  in. 
wide,  and    loft.  4  in.  high,  deducting  a  fire-place  4  11.  4  in.  by  4  ft., 
ami  2  windows  each  G  ft.  by  3  ft.  2  in.?  Ans.  73.^  sq.  yd. 

il.  Find  the  cost  of  glazing  the  windows  of  a  house  of  3  stories,  at 
20 ct.  a  s<|.  t't.  Knch  g'orv  has  4  windows,  3  ft.  10  it1,  wide;  those  in 
the  1st  story  are  7  ft.  8  in.  high ;  those  in  the  2d,  G  ft.  10  in.  \n\'\\\  in 
the  3d,  o  ft.  3  in.  high.  Ant.  $GO.OGjj 


PLASTERERS',  PAINTERS',  AND   PAVERS'    WORK.      353 


10.  The  cost  of  lining  a  tank  2  ft.  10  in.  long,  2  ft.  G  in.  deep,  2  ft. 
wi'le,  with  zinc,  at  10  Ib.  to  the  sq.  ft.,  and  G  ct.  a  Ib.      Ant.  $17.00 

11.  Find  the  cost  of  paving  a  court  60  ft.  by  20  ft.  6  in.,  at  7»  ct.  a 
sq.  yd.  Ans.  $85.12 

12.  Of  paving  a  court  150  ft.  square ;  a  walk  10  ft.  wide  around  the 
whole  being  paved  with  flag-stones  at  64  ct.  a  sq.  yd.,  and  the  rest  at 
81  A  ct.  a  sq.  yd.  Ans.  $927.60 

A  RT.  430.  A  Circle  is  a  figure  bounded  by  a  circum ferencet 
which  is  everv-where  equally  distant  from  a  center  within. 

The  diameter  is  any  line  passing  through  the  center  and  terminated 
on  each  side  by  the  circumference;  the  radius  is  half  the  diameter, 
being  the  distance  from  the  center  to  the  circumference. 

ART.  437.     To  find  the  circumference  from  the  diameter. 
RULE.— Multiply  the  diameter  by  3.14159205,  or  f^-f. 
NOTE. — f  ]  3  may  be  used  for  3.14159265,  being  nearly  the  same. 

1.  What  are  the  circumferences  whose  diameters  are  1G;  22 } ;  72.16; 
and  452  yd.?  Ana.  50.265482;  69.900436;  226.6973;  1420yd. 

ART.  438.     To  find  the  diameter  from  the  circumference. 
RULE. — Divide  the  circumference  by  3.141592G5,  or  f  *§. 

1.  What  are  the  diameters  whose  circumferences  are  5G,  182i, 
316.24  and  639  ft.  ?  Ans.  17.82539,  58.09,  100.66232  and  203.4  ft. 

ART.  439.    To  find  the  area  of  a  circle  from  the  diameter. 
RULE. — Square    half  of  Ike    diameter    and    multiply    it    by 
3.14159205  (or  ff£). 

To  find  the  area  of  a  circle  from  the  circumference. 
RULE. — Square    half   the    circumference    and    divide    it    6j 
3. 14159205  (or  f ff). 

To  find  the  area  of  a  circle  from  the  circum.  and  diameter. 

RULE. — Multiply  the  circumference  by  \  of  the  diameter. 

NOTE. — The  last  rule  can  be  shown  to  be  identical  with  the  others 
by  Art.  437;  to  prove  it,  consider  a  circle  made  up  of  small  triangles, 
having  their  bases  on  the  circumference  and  their  vertices  at  the 
cmter.  The  area  of  each  triangle  is  equal  to  the  base  multiplied  by 
half  its  altitude  (the  radius).  (Art.  432.)  All  of  them,  or  the  circle, 
will  be  equal  to  the  sum  of  their  bases,  that  is,  the  circumference, 
multiplied  by  half  the  radius,  or,  which  is  the  same,  by  one-fourth  of 
the  diameter. 

1.  Find  the  arensof  the  circles  with  diameters  10ft.;  2  ft.  5  in.;  13yd 
I  ft.  Ans.  78.54  sq.  ft. ;  660.52  sq.  in. ;  139  sq.  yd.  5.637  sq  ft. 

2.  Whose  circumferences  are  46  ft.;  7  ft.  3  in.;  6yd.  1  ft.  4  in. 

Ans.  168.386 sq.  ft.;  4  sq.  ft.  26.322  sq.  in.;  29.7443  sq.  ft 
3    Circum.,  47.124  ft.,  diameter  16ft,  Ans.  176. 715  sq.  ft 

30 


354  RAY'S   HIGHER   ARITHMETIC. 


4.   Find  the  area  of  a  ring  whose  breadth  is  2  in.,  and  diameter  in- 
aide,  9  in.  ,4ns.  »Vj.ll6-f-  sq.  in. 

ART.  440.  To  find  the  diameter  or  circumference  from  the  area. 

R  r  t,  E.  —  Divide     the    area,    expressed     in    square    units,    l/y 
3  14  I59'«JG5  (or  ?ff )/  the  square  root  of  the  quotient  will  be  the 
radius;  twice  the  radius  will  be  the  diameter;  and  3.14151)205 
or  y'jj)  times  the  diameter  will  be  the  circumference. 

1.  Find  the  diameter  and  circumference  of  a  circular  field  con- 
taining 10  A.  Ans.  D.  45.14  rd.,  circ.  141.8  rd 

2.  Of  a  circle  containing  8sq.  ft.  HGsq.  in. 

Ans.  I).  40.18  in.,  circ.  12G.23  in. 


MENSURATION    OF   SOLIDS. 

ART.  441.  A  Prism  is  a  solid  having  two  equal  and  parallel 
bases,  and  whose  longitudinal  faces  are  parallelograms. 

The  bases  may  be  triangles,  quadrilaterals,  or  figures  with  any 
number  of  sides,  and  the  prism  is  called  triangular,  quadrangular,  &c., 
accordingly. 

If  the  bases  are  parallelograms,  the  solid  is  called  a  parallelopiped ; 
if  they  are  circles,  it  is  called  a  cylinder. 

The  altitude  of  any  prism  or  cylinder  is  the  perpendicular  between 
the  levels  of  the  bases. 

ART.  442.    To  find  the  solidity  of  any  prism  or  cylinder. 

Rri.E. — find  the  area  of  one  base  in  square  iuii(.i.  and  multi- 
ply it  by  the  altitude  expressed  in  linear  unitx  of  the.  some  name; 
the  product  will  be  the  solidity  in  cubic  units  of  that  name. 

NOTE. — The  area  of  the  base,  whether  it  be  a  triangle,  parallelo- 
gram, or  circle,  is  found  by  one  of  the  previous  rules. 

1.  What  is  the  solidity  of  a  prism  whose  bases  are  squares  9  in.  on 
a  side,  and  whose  altiliub  is  1  ft.  Tin.?  Ans.  153'.)  cu.  in. 

2.  Whose  altitude  is  (>~i  ft.,  and  whose  bases  are  parallelograms  2ft. 
10 in.  long  by  1  ft.  Sin.  w'ide?  Ans.  30 cu.  i't.  1'JOOcu.  in. 

3.  Whose  altitude  is  7  in.  and  whose  base  is  a  triangle  with  a  base 
of  8  in.  and  an  altitude  of  1  ft.?  Ans.  330  cu.  in. 

4.  Whose  altitude  is  4ft.  4  in.  and  whose  base  is  a  triangle  with 
Bides  of  2,  1\  and  3  ft.?  An*.  10.75  cu.  ft. 

5.  What  is  the  solidity  of  a  cylinder  whose  altitude  is  10.1  in.  and 
the  diameter  of  whose  base  is  5  in.?  Ann.  200.1(57  cu.  in. 

ART.  443.  A  Pyramid  is  a  solid  with  a  single  base,  and 
tapering  regularly  to  a  point  called  the  vertex. 

The  base  may  be  a  triangle,  quadrilateral,  &c.,  and  the  pyramid  it 
Balled  triangular,  quadrangular,  &c.,  accordingly. 

If  the  base  is  a  circle,  the  solid  is  called  a  cone. 

The  altitude  of  a  pyramid  or  cone  is  the  perpendicular  from  the 
rertex  to  the  base  or  the  level  of  the  base. 


MENSURAIION    OF   SOLH.S.  355 

ART.  444.     To  find  the  solidity  of  any  pyramid  or  cone. 

RULE. — Find  the  area  of  (he  base  in  square  unity;  multiply  it 
by  one-third  of  the  altitude  expressed  in  linear  units  of  the  same 
name;  the  product  will  be  the  solidity  in  cubic  units  of  that  name. 

1.  Find  the  solidity  of  a  pyramid  whose  altitude  is  1  ft.  2  in.,  aiid 
whose  base  is  a  square  4^  in.  to  a  side.  Ana.  94 .3  cu.  in. 

2.  Whose  altitude  is  15.24  in.  and  whose  base  is  a  triangle  having 
each  side  1  ft.  Ans.  310.70  cu.  in. 

3.  Whose  altitude  is  69  ft.  and  whose  base  is  a  parallelogram  34  ft. 
long  by  '20  ft.  wide.  Ans.  '20332  cu.  ft. 

4.  The  solidity  of  a  cone  whose  altitude  is  5ft.  3  in.  and  the  di- 
ameter of  whose  base  is  2  ft.  1  in.  Ans  5  cu.  ft.  1008.35  cu.  in. 

ART.  445.  A  right  prism  or  right  cylinder  stands  perpen- 
dicular to  its  bases,  the  altitude  being  equal  to  the  length. 

A  right  pyramid  is  one  whose  vertex  is  equally  distant  from  the 
angular  points  of  the  baseband  the  sides  of  whose  base  are  all  equal. 

A  right  cone  is  one  whose  vertex  is  equally  distant  from  all  points 
in  the  circumference  of  the  base. 

In  a  right  pyramid  and  right  cone  the  altitude  falls  at  the  center 
of  the  base. 

ART.  446.    The  surface  of  a  solid  is  its  outside  or  visible  part. 
The  roni'fi  surface  is  nil  the  surface  but  the  base  or  bases. 
The  nl'inl  fiiy/il  of  a  right  pyramid  or  right  cone  is  the  shortest  dis- 
tance from  the  vertex  to  the  boundary  of  the  base. 

ART.  447.  To  find  the  convex  surface  of  a  right  prism  or  right 
cylinder. 

RCI.E. — Multiply  fhe  boundary  of  the  base  by  (he  altitmle,  after 
expressing  them  both  in  the  name  denomination  ;  (he  proa" net  will 
be  the  convex  surface  in  square  units  of  the  same  name. 

NOTES. — 1.  To  get  the  whole  surface,  add  in  the  areas  of  the  two 
bases. 

2.  To  prove  the  rule,  consider  that  the  convex  surface  of  a  right 
prism  or  right  cylinder  when  rolled  out  on  a  plane,  becomes  a  rec- 
tangle whose  adjacent  sides  are  the  altitude  of  the  solid  and  the  boun- 
dary of  its  base. 

1.  Find  the  convex  surface  of  a  right  prism  with  altitude  11  j  in. 
and  sides  of  base,  6.J,  OA,  8j,  10.1,  9  in.  Am.  450  «q.  in. 

2.  Of  a  right  cylinder  whose  altitude  is  l|  ft.  and  the  diameter  of 
whose  base  is  1  ft.  2i  in.  Ans.  0  sq.  ft.  92.0  sq.  in. 

3.  Find  the  whole  surface  of  a  right  triangular  prism,  the  sides  of 
the  base  00,  80.  and  lUO  ft.;  aliitude  90  ft.  Ann.  204IHJ  sq.  ft. 

4.  The  whole  surface  of  a  cylinder;  altitude  28  ft.;  circumfrrenct 
of  the  base  19ft.  A/,.i.   689.455  sq.  ft. 

ART.  448.  To  find  the  convex  surface  ol  a  right  pyramid  01 
right  cone. 


350  RAY'S   HIGHER  ARITHMETIC. 

RULE. — Multiply  the  boundary  of  tite  base  by  half  the  slant 
kit/lit,  after  expressing  them  in  (he  same  denomination ;  the  pro- 
duct will  bt  the  convex  surface  in  square  units  of  the  same  name, 

NOTKS. — 1.  To  get  the  whole  surface,  add  in  the  area  of  the  base. 

2.  To  prove  the  rule,  consider  the  convex  surface  as  made  up  of 
triangles  reaching  from  the  vertex  to  the  boundary  of  the  base,  their 
altitude  being  the  slant  hight  of  the  solid,  and  their  bases  making 
I  lie  boundary  of  its  base. 

1.  Find  the  convex  surface  and  whole  surface  of  a  right  pyramil 
whose  slant  hight  is  225  ft. ;  the  base  040  ft.  square. 

Am.  Conv.  surf.  288000  sq.  ft. ;  whole  surf.  G97COO  sq.  ft. 

2.  Of  a  right  cone  whose  slant  hight  is  66 ft.  Sin. ;  radius  of  the 
base  4  ft.  2  in.  Ans.  12-3663.706  sq.  in. ;  133517.6876  sq.  in. 

ART.  449.  A  Sphere  or  Globe  is  a  solid  bounded  by  a  curved 
surface,  which  is  at  all  points  equally  distant  from  its  center. 

The  diameter  of  a  sphere  is  any  line  passing  through  the  center  and 
terminated  both  ways  by  the  surface. 

The  radius  is  half  the  diameter,  being  the  distance  from  the  center 
to  the  surface. 

ART.  450.  To  find  the  surface  of  a  sphere  from  its  diameter. 

RULE. — Multiply  the  square  of  the  diameter  by  3.14159265  (or 
ffl)i  the  product  will  be  the  surface  in  square  units. 

NOTE. — The  surface  of  any  sphere  is  just  equal  to  the  area  of  four 
circles  having  the  same  diameter  as  the  sphere. 

1.  What  are  the  surfaces  of  two  spheres  whose  diameters  are  27  ft. 
and  10  in.  ?  Ans.  2290.221-}-  sq.  ft.,  and  314.10  sq.  in. 

ART.  451.  To  find  the  solidity  of  a  sphere  from  its  diameter. 
RULE. — Multiply  the  cube  of  the  diameter  by  3.14159265  (or 
rt;j)/  t°ke  Q  of  the  product :  this  will  be  the  solidity  in  cubic  units. 
To  find  the  solidity  of  a  sphere  from  its  surface  and  diameter, 

RITLE. — Multiply  the  surface  expressed  in  square  units  by  g  of 
the  diameter  expressed  in  linear  units  of  the  same  name;  the  pro- 
duct li'ill  be  the  solidity  in  cubic  units  of  that  name. 

NOTES. — 1.  If  the  surface  only  is  given,  divide  it  by  3.14159265 
(or  yflj) ;  the  square  root  of  the  quotient  will  be  the  diameter:  then 
apply  the  rule. 

2.  To  prove  the  second  rule,  which  is  identical  with  the  first,  con- 
sider the  sphere  to  consist  of  small  pyramids  having  their  vertices 
»t  the  center  and  their  bases  on  the  surface,  the  altitude  of  each 
being  the  radius  <>f  the  sphere. 

1.  Find  the  solidity  of  a  sphere  whose  diameter  is  0  mi.,  and  sur 
face,  113.007335  sq.  mi.  Ans.  113.097335  cu.  mi. 

2.  Of  a  sphere  whose  diameter  is  4  ft.  Ans.  33.5103  cu.  ft. 

3.  Of  a  sphere  whose  surface  is  40115sq.  mi.  Ans.  755499 g  cu.  mi. 


MASONS'    AND   BRICKLAYERS'    WORK.  357 

ART.  452.  A  Fmslum  of  a  pyramid  or  cone  is  what  remains 
after  cutting  off  its  top,  making  the  upper  base  parallel  to  the  lower 

The  altitude  of  a  frustum  is  the  perpendicular  or  shortest  distance 
between  the  levels  of  the  bases. 

The  slant  hight  of  a  frustum  of  a  right  pyramid  or  of  a  right  cone, 
is  the  shortest  distance  between  the  boundaries  of  its  bases,  measured 
on  its  convex  surface. 

ART.  453.  To  find  the  convex  surface  of  a  frustum  of  a 
right  pyramid  or  of  a  right  cone. 

RULE. — Take  half  the  sum  of  the  boundaries  of  ifs  two  bases, 
and  multiply  it  by  the  slant  hight,  after  expressing  them  in  the, 
same  denomination,:  the  product  will  be  the  convex  surface  in 
square  units  of  the  same  name. 

NOTES. — 1.  To  get  the  whole  surface,  add  in  the  areas  of  the  bases. 
2.  The  reason  of  this  rule  is  seen  from  that  of  the  trapezoid. 

1.  Find  the  convex  surface  of  a  frustum  of  a  pyramid  with  slant 
hight,  3  j  in.;  lower  base,  4  in.  square;  upper  base.  2A  in.  square. 

Ant.  43 1  sq.  in. 

2.  The  convex  surface  and  whole  surface  of  the  frustum  of  a  cone; 
the  diameters  of  the  bases,  7  in.  and  3  in.;  the  slant  hight,  5  in. 

Ans.  Conv.  surf.  78.5398  sq.  in.,  whole  surf.  124.0929  sq.  in. 

ART.  454.  To  find  the  solidity  of  a  frustum  of  a  right  pyra- 
mid or  frustum  of  a  right  cone. 

RULE. — Find  the  areas  of  the  two  bases,  also  of  a  mean  base 
equal  to  the  square  root  of  the  product  of  the  other  two ;  take  the 
sum  of  the  three,  and  multiply  it  by  \  of  the  altitude  ;  the  product 
will  be  the  solidity  in  cubic  wills. 

1.  Find  the  'Kiidity  of  a  frustum  of  a  pj'ramid  whose  altitude  is 
1  ft.  4^  in.;  lower  base,  10$  in.  square;  upper,  4.J  in  square. 

Ans.  1)74 A ||  cu.  in. 

2.  Of  a  frustum  of  a  cone;  the  diameters  of  the  buses  being  18 in. 
and  10  in. ;  the  altitude  10  in.  Ans.  2530.03  cu.  in. 


MASONS'  AND  BRICKLAYERS'  WORK. 

ART.  455.  Masons'  \vork  is  sometimes  measured  by  the  cubic 
foot,  and  sometimes  by  the  perch,  \vhieh  is  1(5 A  ft.  long,  1 A  ft.  wide, 
»ni  1  ft.  deep,  and  contains  1GA  X  H  X  1  =  '<M|  cu.  ft.,  or  25 
tu.  ft.,  nearly. 

ART.  456.    To  find  the  No.  of  perches  in  a  pieco  of  masonry. 

RIM.E.-  -Find  the,  solidify  nf  the  wall  in  cubic  feel  by  the  rules 
given  for  mensuration  of  solids,  and  dicide  it  by  "2A\. 

NOTE. — Brick  work  is  generally  estimated  by  the  thousand  bricks 
the  usual  size  being  8  in.  long,  4  in.  wide,  and  2  in.  thick.  When  bricks 
»re  laid  in  mortar,  an  allowance  of  -j'jj  is  made  for  the  mortar. 


358  RAY'S   HIGHER   ARITHMETIC. 

1.  How  many  perches  of  25  cu.  ft.,  in  a  pile  of  building  stone  18  ft 
long.  ^<.>  ft.  wide,  and  b  f t.  2  in.  high?         Ans.  37.74  =  :>7|  nearly. 

2.  Finil  the  cost  of  laying  a  wall  20  ft.  long,  7  ft.  9  in.  high,  and  will 
a  mean  breadth  of  2  ft.,  at  75  ct.  a  perch.  Ant.  $'.!.:>'.) 

3.  The  cost  of  a  foundation  wall  1ft.  10  in.  thick,  and  9  ft.  4  in.  high, 
for  a  building  80 ft.  long,  22ft.  5in.  wide  outside,  allowing  for  2  doors 
4  ft.  wide,  at  S2.75  a  perch.  Ann.  $192.'.»8 

4.  The  co«t  of  a  brick  wall  150  ft.  long,  8  ft.  G  in.  high,  1  ft.  4  in.  thick, 
allowing  y0  for  mortar,  at  $7  a  thousand.  Arts.  $2«'J.17 

5.  How  many  bricks  of  ordinary  size  will  build  a  square  chimney 
86  ft.  high,  10  ft.  wide  at  the  bottom  and  4  ft.  at  the  top,  outside,  and 
3ft.  wide  inside  all  the  way  up?  Am.  89801-)- 

SUOOESTION. — Find  the  solidity  of  the  whole  chimney,  then  of  the 
hollow  part ;  the  difference  will  be  the  solid  part  of  the  chimney. 


GAUGING. 

ART.  457.  Gauging  is  finding  the  contents  of  vessels,  in 
bushels,  gallons,  or  barrels. 

To  gauge  any  vessel  in  the  form  of  a  rectangular  solid,  cylinder, 
cone,  frustum  of  a  cone,  &c. 

KIM.E. — Find  the  solidity  of  the  vessel  in  cubic  inches  by  the 
rules  ali-fail it  t/iten  ;  this  dicided  by  2150.42  will  (/ice  Hie  contents 
in  bn.;  by  282  idfl  give  it  in  beer  //«/.;  by  231  will  </ice  if  in  irine 
gal.,  which  may  be  reduced  to  bbl.  by  dividing  them  by  31  3. 

NOTE. — In  applying  the  rule  to  cylinders,  cones,  and  frustums  of 
cones,  instead  of  multiplying  the  square  of  half  the  diameter  by 
3.1415020/3,  and  dividing  it  by  231,  multiply  tht  square  of  the  diameter 
by  .0034,  which  amounts  to  the  same,  and  is  shorter. 

1.  How  many  bushels  in  a  bin  8ft.  3 in.  long,  3ft.  Sin.  high,  and 
2ft,  10  in.  wide?  Ant.  04.18bu. 

2.  How  many  wine  gallons  in  a  bucket  in  the  form  of  a  frustum 
of  a  cone,  the  diameters  at  the  top  and  bottom  being  13  in.  and  10  in., 
and  depth  12  in.?  Ans.  5.4264  gal. 

3.  How  many  barrels  in  a  cylindrical  cistern  11  ft.  6  in.  deep,  and 
7  ft.  8  in.  wide  ?  Aru.  126.0733 

4.  In  a  vat  in  the  form  of  a  frustum  of  a  pyramid,  5  ft.  deep,  10ft. 
f-juare  at  top,  9  ft.  square  at  bottom  ?  Ans.  107.20  bbl. 

6.  In  a  well  3  ft.  wide,  and  12  ft.  deep  ?  Am.  20.1435  bbl. 

A  RT.  458.    To  find  the  contents  in  gallons  of  a  cask  or  barrel 
When  the  staves  are  straight  from  the  bung  to  each  end,  consider 
the  ca><k  as  two  frustums  of  a  cone,  and  calculate  its  contents  by 
last  rule;  but  when  the  staves  are  curved,  use  this 

Rri,E.—  Add  to  the  head  diameter  (inside")  two-thirds  of  tht 
difference  between  the  head  and  bnng  diameters;  but  if  the,  stave* 


TUNNAGE  OF  VESSELS.  359 

art  only  slight!;,'  curved,  add  six-tenths  of  this  difference ;  this  given 
the  mean  diameter;  exprexx  it  in  inches,  syuare  it.  >mil/i/>ly  it  Inj 
the  length  in  inches  and  thin  product  by  .003-1.'  the  product  wit' 
be  Ike  contents  in  wine  gallons. 

NOTE. — After  finding  the  mean  diameter,  the  contents  are  foun» 
as  if  the  cask  were  a  cylinder. 

1.  Fiml  the  number  of  gallons  in  a  cask  of  beer  whose  staves  are 
straight  from  bung  to  head,  the  length  being  20  in.,  the  hung  diameter 
10  in.,  and  head  diameter  13  in.  Ans.  10. -K  gal. 

2.  In  a  barrel  of  whisky,  with  staves  slightly  curved,  length  '2  ft. 
10  in.,  bung  diameter  1  ft.  9  in.,  head  1  ft.  0  in.  Ans.  45.42  go  I. 

3.  In  a  cask  of  wine,  with  curved  staves,  length  5  ft.  4  in.,  bung 
diameter  3  ft.  6  in.,  head  diameter  3  fL  Ans.  348.1G  gal. 


TUNNAGE    OF    VESSELS. 
CARPENTERS'   RULE. 

AUT.  459.  For  sin  file-decked  vessels,  multiply  together  the 
length  of  the  keel,  the  breadth  at  the  main  beams,  and  the  depth  of 
the  /mill,  all  in  fuel,  and  divide  the  product  by  (.)5. 

If  the  vessel  in  double-Jerked,  proceed  ax  for  single-decked  ves- 
W/M.  exrepl  that  the  di-fith  of  the  hold  is  not  measured,  but  assumed 
tt>  be  half  the  breadth  at  the  main  beam. 

GOVERNMENT  RULE. 

If  the  vessel  is  double-decked,  measure  its  length  from  Ilie  fore 
part  of  the  main  xtem,  to  the  after  part  of  the  stern-post,  abort 
tlif.  upper  deck ;  measure  the  breadth  at  the  broadest  purl  ab»ce 
the  main  vales,  half  of  which  breadth  shall  be  accounted  the 
depth;  then  deduct  from  the  length  three-Jifths  of  the  breadth, 
nnif/ifily  the  remainder  In/  the  breadth,  and  the  product  by  the 
depth,  and  divide  the  result  by  95  for  the  tunnage. 

If  the  vessel  is  single-decked,  proceed  as  with  double-decked 
rc.v.ve/.v.  except  that  the  depth,  instead  of  being  assumed  half  of  the 
breadth,  should  be  measured  from  the  under  side  of  the  deck-plank 
to  the  ceiling  of  the  hold. 

AllT.  460.    To  find  the  tunnage  more  accurately,  use  this 

RULE. — Find  the  plumb  or  perpendicular  hight,  from  the  lowest 
point  of  1 he  deck  to  the  water-line ;  take  a  mrun  breadth  between 
the  breadth  on  deck  and  the  breadth  at  the  water  line  ;  also  a 
mean  length  of  ludl  between  the  length  on  deck  and  the  length  at 
the  water-line;  express  these  dimensions  in  feel ;  wnllip/y  them 
togtthcr,  and  divide  the  product  by  .'J5.N4  Jor  the  tunnage. 

NOTES. — 1.  The  water  line  is  the  line  made  on  the  outside  of  the 
hull,  by  the  surface  of  the  water,  when  the  vessel  floats  without  any 
load. 


3GO  RAY'S  HIGHER   ARITHMETIC. 

2.  The  product,  of  the  menu  length,  breadth,  and  depth  to  the  •water- 
line,  gives  the  number  of  cubic  feet  displaced,  \>y  pulling  »  lull  lend 
in  the  vessel;  this,  multiplied  by  (i'J.l,  givt-s  the  weight  of  that  water 
in  lb.,  aud  (his  divided  by  2240  becomes  tuns;  but,  instead  of  mul- 
tiplying by  02.',,  and  dividing  by  2240,  get  the  tuns  at  ouce  by  di- 
viding by  35.84,  since  ^-  =- — -.  This  water  is  of  the  same  weight 
as  the  load  which  displaces  it;  that  is,  the  tunnage  of  the  vessel. 

1.  What  are  the  carpenters'  and  government  tunnage  of  a  single 
decked  vessel,  whose  length  of  keel  is  84  ft.,  breadth  of  beam  2K  ft., 
and  depth  of  hold  9  ft.?         Ans.  Carp.  223  T.  nearly;  Gov.  178J  T. 

2.  Also  a  double-decked  vessel,   150  ft.  keel  and   30]  ft.  breadth 
of  beam?  Ana.  Carp.  1037.^  T.  nearly;  Gov.  8*7  T. 

3.  Also,  by  the  3d  rule,  of  a  vessel,  mean  length,  93  ft.  4  in.,  mean 
breadth,  26  ft.  7  in.,  depth  to  the  water  line,  4  ft.  3  in.?    Ans.  294  T. 

4.  How  many  bushels  of  coal  (80  lb.)  can  be  shipped  on  a  boat 
78  ft.  long,  22  ft.  wide,  and  5ft.  to  the  water  line,  leaving  1  ft.  Sin. 
above  the  water?  Ans.  4468|  bu. 


XLI.    MECHANICAL  POWERS. 

ART.  461.  The  Mechanical  Powers  are  six,  viz.:  the  Lever, 
Wheel  and  Axle,  Inclined  Plane,  Screw,  Pulley,  and  Wedge. 

All  machines  contain  one  or  more  of  these  powers. 

The  power  in  a  machine  is  the  force  applied,  as  steam,  horses, 
weights,  hand-power,  &c. 

The  weight  is  the  obstacle  overcome. 

One  of  the  chief  obstacles  to  the  working  of  machinery  is  friction; 
that  is.  the  rubbing  and  interference  of  the  parts  of  the  machine 
itself:  this  is  generally  estimated  at  3  of  the  power  applied. 

ART.  462.  The  power  and  weight  in  a  machine  are  to  each 
other  inversely  as  their  velocities  ;  hence, 

TO   COMPARE   THE   POWER   AND   WEIGHT   IN   A   MACHINE, 

ROI.E.—  The.  power  :  the  weight  :  :  the  distance  the  weight 
moves  in  the  direction  of  resistance  :  the  distance  the  power 
passes  through  in  the  same  time. 

NOTE.-- Any  three  ferms  being  given  in  this  proportion,  the  fourth 
may  be  found  by  Art.  249. 

THE   LEVER 

ART.  463.  Is  a  bar  movable  about  a  fixed  point  called  a 
fulcrum  ;  there  are  three  kinds: 

1st..  When  the  power  nnd  weight  are  on  different  sides  of  the  fulcrum 


MECHANICAL  POWERS. 


2d.  When  the  power 
ind  weight  aro  on 
die  same  side  of  the 
fulcrum,  the  power 
being  nearer  the  ful- 
crum. 

3d.    When    the    power 


and    the 

weight  are  on  the  same  side  of 
the  fulcrum,  the  weight  being  nearer 
the  fulcrum. 


ART.  464.     To  compare  the  power  and  weight  in  any  lever. 

RULE. —  The  power  :  the  weight  :  :  Vie  distance  of  the  weight 
from  the  fulcrum  :  the  distance  of  the  power  from  the  fulcrum. 

NOTE. — This  rule  is  the  same  in  fact  as  the  general  rule  (Art.  4G2); 
for  the  distances  passed  over  by  the  power  and  weight  in  the  same 
time,  depend  on  their  distances  from  the  fulcrum. 

1.  What  weight  10  in.  from  the  fulcrum,  is  balanced  by  a  power  of 
40  lb.,  4  ft.  8  in.  from  the  fulcrum  ?  Ans.  2:24  Ib. 

2.  What  power,  6  ft.  from  the  fulcrum,  will  sustain  a  weight  of 
1080  lb.,  1  ft.  4  in.  from  the  fulcrum  ?  -4ns.  240  lb. 

3.  How  far  from  the  fulcrum  must  a  person  weighing  160  lb.  stand, 
to  balance  a  weight  of  1200  lb.,  2  ft.  8  in.  from  the  fulcrum?  Ans.  20  ft. 


THE  WHEEL  AND  AXLE 


ART.  465.  Is  a  sort  of  lever.  The 
radius  of  the  wheel  and  the  radius  of 
the  axle  are  parts  of  the  lever,  and  the 
center  of  the  axle  is  the  fulcrum. 

The  power  nets  at  the  circumference  of 
the  wheel,  and  the  weight  at  the  circum- 
ference of  the  axle,  by  means  of  a  rope. 
The  distances  passed  over  by  the  power 
and  weight  in  the  same  time,  will  be  the 
circumferences  of  the  wheel  and  of  the 
axle  ;  hence, 


RULE. —  The  power  :  the  weight  :  :  the.  circumferenct  (or  diame- 
ter) of  the  axle  :  the  circumference  (or  diameter  of  the  wheel). 

NOTE. — The  diameters  may  be  used  instead  of  the  circumferences 
if  desired,  since  they  vary  as  the  circumferences. 

1.  The  diameter  of  the  wheel  is  G  ft.,  of  the  axle,  8  in. ;  the  weight 
is  720  lb. :  find  the  power,  allowing  j  for  friction.         Am.  106 'j  lb. 

2.  The  diameter  of  the  wheel  is  0  ft.  4  in.,  of  the  axle  1  ft.  2  in.;  the 
power  is  18-3  lb. :  find  the  weight,  friction  i.  Ans.  1184  lb. 

3.  The  diameter  of  the  axle  is  10  in.,  the  power,  140  lb.,  the  weigh*, 
1750  lb  :  find  the  diameter  of  the  wheel.  Ar&.  10  f»   R  in. 

31 


362 


RAY'S   HIGHER   ARITHMETIC. 


THE   INCLINED   PLANE 

ART.  466.  Ir  used  to  raise  or  lower  heavy  bodies.  The  power 
uiu>t  f:il I  a  distance  equal  to  the  length  of  the  plane,  in  order  to 
raise  the  weight  a  distance  equal  to  its  hight ;  hence, 

Hi; I.E. —  The  power  :  ike  weight  :  :  the  liujht  of  the  plane  :  the 
length  r<f  ike  plane. 

1.  What  power  -will  roll  a  bbl.  of  flour  (19Glb.)  up  an  inclined 
plane  7  ft.  G  in.  *ong,  into  a  wagon  3  ft.  4  in.  high  ?         Am.  87  £  Ib. 

2.  W hat  weight  can  a  man  capable  of  lifting  300  Ib.  roll  up  an 
inclined  plane  8  ft.  long,  and  5  ft.  4  in.  high  ?  Ans.  450  Ib. 

THE   SCREW- 

ART.  467.  May  be  considered  an  inclined  plane,  called  the 
thread,  wrapped  round  a  cylinder. 

The  power  is  generally  applied  by  a  lever  fixed  firmly  into  the 
head  of  the  cylinder.  The  power  describes  the  circumference  of  a 
circle  whose  radius  is  the  lever,  while  the  weight  is  moved  the  dis- 
tance between  two  contiguous  threads ;  hence, 

RULE. — The  power  :  the  icenJit  :  :  the  distance  bcticeen  two  con- 
tiguous thready  :  f  y  3  times  twice  the  length  of  the  lever. 

NOTK. — The  last  term  is  the  circumference  described  by  the  power. 

1.  What  weight  will  be  moved  by  a  power  of  500  Ib.  acting  at  the 
end  of  a  lever  15  ft.  long,  by  means  of  a  screw  whose  threads  are  1~,  in. 


apart,  allowing  3  for  friction? 


Ans.  251327  Ib. 


2.    What  power  will  lift  28400  Ib.  by  a  screw  with  threads  |  of  an 
inch  apart,  and  a  lever  6ft.  long?          Ans.  47y':jlb. 

THE   PULLEY 

ART.  468.  Is  a  grooved  wheel,  movable  about 
its  center,  the  power  and  weight  being  connected 
by  a  cord  running  in  the  groove. 

The  single  pulley  affords  no  advantage  except  in 
the  direction  of  applying  the  power. 

In  a  system  of  pulleys,  observe  that  for  every 
movable  pulley  around  which  the  cord  runs,  the 
power  passes  over  twice  the  distance  the  weight  is 
lifted;  hence, 

RULE. — The  power  :  the  weight  :  :  1  :  twice  the 
number  of  movable  pulleys. 

1.  The  power  is  25  Ib.,  the  number  of  movable 
pulleys  8  :  what  is  the  weight  ?  Ans.  400 Ib. 

2.  The  weight  is  loOOlb.,  the  number  of  movable 
•ulleys  6  :  what  is  the  power?  Ans.  125  Ib. 

NOTE. — The  Wedge  is  a  double  inclined  plane, 
and  the  power  :  the  weight  :  :  the  head  of  the 
wedge  :  its  length;  but  the  friction  is  so  great  in 
this  machine,  and  the  power  is  so  much  influenced 
by  other  considerations,  that  it,  is  very  difficult  to 
obtain  satisfactory  practical  results  by  any  method 
of  calculation,  and  hence  no  examples  are  given. 


PROMISCUOUS  QUESTIONS.  3G3 


ART.  469.    ILLUSTRATIONS  OF  MECHANICAL  POWERS. 

PROBLEM  1. — To  find  the  pressure  of  the  steam  in  pounds,  ou  each 
iquare  inch  of  the  surface  of  the  boiler  of  a  steam  engine. 

lli'LK. — Multiply  the  iceigld  at  the  end  of  the  steam  guagc  It/  i/s 
distance  from  the  fulcrum,  and  divide  the  product  by  the  distance 
of  the  valve  front  the  fulcrum;  this  quotient  divided  by  f  '\$  times 
tlit  square  of  half  the  diameter  of  the  voice,  gives  the  pressure  on 
ta:h  square  inch  of  the  boiler. 

NOTE. — The  distances  and  diameter  of  the  valve  must  be  in  inches. 

1.  The  weight  on  the  guage  is  58  lb.,  and  is  1  ft.  10  in.  from  the 
fulcrum;  the  valve  is  4 in.  from  the  fulcrum,  and  its  diameter  is  3 in.: 
what  is  the  pressure  ?  Ans.  45  lb.  to  the  sq.  in. 

?.  The  weight  is  3Glb.,  its  distance  from  the  fulcrum  2ft.  4  in.; 
the  valve  is  4i  in.  in  diameter,  and  is  3  in.  from  the  fulcrum  :  what 
is  the  pressure  ?  Ans.  21  lb.  to  the  sq.  in. 

PROB.  2. — To  find  the  horse  power  of  a  steam  engine. 

RULE. — Multiply  the  pressure  in  lb.  on  the  square  inch  of  the 
boiler  by  jjjj  times  the  square  of  half  the  diameter  of  the  cylinder 
in  inches,  and  this  product  by  twice  the  length  of  the  cylinder  in 
feet,  and  this  product  by  the  number  of  double  strokes  made  by 
the  piston  in  a  minute,  and  divide  this  by  4-1000  ;  the  quotient  is 
the  horse-power  of  the  engine. 

NOTE. — This  rule  makes  no  allowance  for  the  cooling  of  the  steam 
in  passing  from  the  boiler  to  the  cylinder,  by  which  its  elastic  force 
is  diminished.  In  high-pressure  engines  this  is  considerable,  and 
in  some  more  than  others.  An  allowance  may  be  made  for  this, 
l>;ised*on  experience,  or  the  pressure  of  the  steam  in  the  cylinder 
may  be  determined  by  its  temperature. 

3.  If  the  pressure  is  34  lb.  per  sq.  in.,  the  diameter  of  the  cylinder 
2  ft.  G  in.,  its  length  7  ft.,  and  the  piston  makes  18  double-strokes  per 
minute,  what  is  the  power  of  the  engine?      Ans.  138  horse-power. 

4.  If  the  pressure  is  44  lb.  per  sq.  in.,  the  diameter  of  the  cylinder 
1  ft.  8  in.,  its  length  G  ft.,  and  the  paddle-wheel  makes  20  revolutions 
a  minute,  what  is  the  power  of  the  engine?      Ans.  75  horse-power. 

To  TEACHERS. — The  Student  is  supposed  to  have  learned  the 
iemonstrations  of  the  Rules  in  series  and  subsequent  subjects,  from 
"Raj's  Arithmetic,  Third  Book":  hence  they  are  omitted  here. 


364  RAY'S  HIGHER  ARITHMETIC. 


XLII.   PROMISCUOUS   EXERCISES. 
ART.  470.     FIFTY  EXAMPLES  TO  BE  ANALYZED. 

1.  If  I  gain  ]  ct.  apiece  by  selling  eggs  at  7ct.  a  dozen,  how  much 
apiece  will  I  gain  by  selling  them  at  U  ct.  a  dozen?        Am.  -f\2  ct- 

2.  If  I  gain  i  ct.  apiece  by  selling  sft»plcs  at  3  for  a  dime,  how 
uch  apiece  would  I  lose  by  selling  them  4  fora  dime?    Ans.  \  ct. 

3.  If  I  sell  potatoes  at  37^  ct.  per  bu.,  my  gain  is  only  f  of  what 
it  would  be,  if  I  charged  45  ct.  per.  bu. :   what  did  they  cost  me? 

Ans.  26  j  ct.  per  bu. 

.     4.  If  I  sell  my  oranges  for  Go  ct.,  I  gain  §  ct.  apiece  more  than  if 
I  sold  them  for  50  ct. :  how  many  oranges  have  I  ?  Ans.  40. 

5.  If  I  sell  my  pears  at  5  ct.  a  dozen,  I  lose  1G  ct.;  if  I  sell  them  at 
8ct.  a  dozen,  I  gain  11  ct.:  how  many  have  I,  and  what  did  they 
cost  me  ?  Ans.  9  dozen  at  Gg  ct.  per  dozen. 

6.  If  I  sell  eggs  at  G  ct.  per  dozen,  I  lose  f  ct,  apiece;  how   much 
per  dozen  must  1  charge,  to  gain  -ij  ct.  apiece?  Ans.  22  ct. 

7.  £  of  a  dime  is  what  part  of  3ct.  ?  Ans.  y, 

8.  If  I  lose  |  of  my  money  and  spend  |  of  the  remainder,   v.-hat 
j.'irt  have  1  left  '.'  Ans.  %'g 

9.  A's  land  is  ^  less  in  quantity  than  B's,  but  •%•§  better  in  quality: 
how  do  their  farms  compare  in  value?  Ans.  A's=y9g  of  B's. 

10.  If  |  of  A's  money  equals  |  of  B's,  what  part  of  B's  equals  £  of 
A's?  Ans.  J 

11.  I  gave  A,  T54  of  my  money,  and  B,  T7j  of  the  remainder:   who 
got  the  most,  and  what  part?          Ans.  B  got  -$s  of  it  more  than  A. 

12.  A  is  f  older  than  B,  and  B,  §  older  than  C:  how  many  times 
is  A  as  old  as  C?  Ans.  1\ 

13.  \  of  my  money  equals    4  of  yours;    if    we  put   our  money 
together,  what  part  of  the  whole  will  1  own?  Am.  T'yj 

14.  How  many  thirds  in  A  ?  Ans.  1  A 

15.  Reduce  4  to  thirds;   g  to  ninths;  'and  f  to  a  fraction,  whose 
numerator  shall  be  8.  Ans.  2j£  thirds;   7r,  ninths;   -^8.,f 

16.  What  fraction  is  as  much  larger  than  i,  as  f  is  less  than  •'-'  ? 

Ans.  ji 

17.  After  paying  out  -{  and  i  of  my  money,  I  had  left  $8  more  than 
I  had  spent :   what  had  I  at  first  ?  Ans. 

18.  In  12 yr.,  I  shall  be  §  of  my  present  age;  how  long  since  was 
I  |  of  my  present  age?  Ans.  8$yr. 

10.  Four  times  §  of  a  number,  is  12  less  than  the  number:  what  ia 
the  n umber?  Ans.  108. 

20.  A  man  left  y\  of  his  property  to  his  wife,  f  of  the  remainder 
to  his  son,  and  the  balance,  §4000,  to  his  daughter:  what  was  the 
estate?  Ans.  $22000. 

21.  I  sold  an  article  for  £  more  than  it  cost  me,  to  A,  who  sold  it 
for  $G,  which  was  §  lesa  than  it  cost  him  :  what  did  it  cost  me  ? 

Ans.  $8. 


PROMISCUOUS   EXERCISES.  365 


22.  A  is  5  older  than  B;  their  father,  who  is^as  old  nsbotb  of  them, 
is  f>0yr.  of  age:  how  old  are  A  and  B?    Arts.  A,  27  i  yr.;   15,  22  \  yr. 

23.  A  pole  was  z  under  water;  the  water  rose  8ft.,  and  then  there 
wiH  as  "inch  under  •water  as  had  been  above  water  before  :  how  long 
is  the  pole?  AHS.  18^  ft. 

24.  A  is  |  as  old  as  B ;  if  he  were  4  yr.  older,  lie  would  be  -^  as 
old  as  B;  how  old  is  each?  Ans.  A,  20yr.;  B,  2Gj  yr. 

25.  A's  money  is  $4  more  than  -|  of  B's,  and  $5  less  than  |  of  B's: 
tow  much  lias  each?  Ans.  A,  $7G;  B,  $108. 

2(5.  -ij  of  A's  age  is  4  of  B's,  and  A  is  3i  yr.  the  older:  how  old  is 
each  ?  '  AM.  A,  3U  yr.;  B,  28  yr. 

27.  If  3  boys  do  a  work  in  7  hr.,  how  long  will  it  take  a  man  who 
works  4-.i  times  as  fast  as  a  boy  ?  Ans.  4rf  hr. 

28.  If  G  men  can  do  a  work  in  5^  days,  how  much  time  would  be 
eafcd,  by  employing  4  more  men  ?  .  Ans.  2  5  days. 

29.  A  man  and  2  boys  do  a  work  in  4  hr. ;  how  long  would  it  take 
the  man  alone,  if  he  worked  equal  to  3  boys  ?  Ans.  63  hr.    ~ 

30.  A  man  and  a  boy  can  mow  a  certain  field  in  8hr.;   if  the  boy 
rests  3|  hr.,  it  takes  them  9|  hr.:  in  what  time  can  each  do  it? 

Ans.  Man,  13 J  hr.;  boy,  20 hr. 

31.  Five  men  were  employed  to  do  a  work  ;   two  of  them  failed  to 
come,  by  which  the  work  was  protracted  4~,  days:  in  what  time  could 
the  5  have  done  it?  Ans.  Gj  days. 

32.  3  men  can  do  a  work  in  5  dnys;   in  what  time  can  2  men  and 
8  boys  do  it,  allowing  4  men  to  work  equal  to  0  boys?    Ans.  4*  da.     \ 

33^  A  man  and  a  15oy  mow  a  10  acre  field;  how  much  more  does 
the  man  mow  than  the  boy,  if  2  men  work  equal  to  5  boys? 

Ans.  4  7  A. 

"""34.  G  men  can  do  a  work  in  4.'  dnys;  after  working  2  days,  how 
many  must  join  them  so  as  to  complete  it  iuXi  ^a-?  Ans.  4  men. 

:'.").  8  men  can  do  a  work  in  G^  days;  after  beginning,  how  soon 
must  they  be  joined  by  2  more,  so  as  10  complete  it  in  5g  days?. 

Ans.  In  2^  days. 

30.  7  men  can  build  a  wall  in  5i  days;  if  10  men  are  employed, 
what  part  of  his  time  can  each  rest,  and  the  work  be  done  in  the 
«ame  time?  Ans.  j\ 

-  J>7.  U  men  cnn  do  a  work  in  8.'  days;  how  many  days  may  3  re- 
main away,  and  yet  finish  the  work  in  the  same  time  by  bringing  5 
more  with  them  ?  Ans.  o^0,- 

38.  10  men  can  dig  a  trench  in  7i  days;  if  4  of  them  arc  absent 
the  first  2.j  days,  how  many  others  must  they  then  bring  with  them, 
to  complete  the  work  in  the  same  time?  Ans.  2. 

89.  At  what  times  between  G  and  7  o'clock  are  the  hour  hand  and 
minute  hand  20min.  apart? 

Ans.  10'j-j  min.  after  G,  and  54T"T  min.  after  G 

40.  At  what  time*  between  4  and  o  o'clock  is  the  minute  hand  as 
far  from  8  as  thn  hour  hand  is  from  3  ? 

Ans.  3274g  after  4;  and  49T'T  min.  after  4. 


360  RAY'S   HIGHER  ARITHMETIC. 

41.  At  what  time  between  5  and  0  o'clock,  is  the  minute  hand  mid- 
tray  bot \vcfn  12  aud  the  hour  liaiul '.'  when  is  the  hour  hand  midwaj 
between  1  and  the  miunte  hand '! 

Aiif.  Ion's  min.  after  5;  ami  3H  niln.  a  ft  or  •">. 

42.  A,  B  and  C  dine  on  8  loaves  of  bread;  A  furnishes  ~>  loaves; 
B,  ;>  loaves;  U  pays  tlie  others  8d.  for  his  share:   how  must  A  ami  B 
divide  the  money?  Ans.  A  takes  7d. ;   i>,  Id. 

4o.  A  beat  makes  15  mi.  an  hour  down  stream,  and  10  mi.  an  hour 
Up  stream:  how  far  can  she  go  and  return,  in  0  hr.?  Ans.  54  mi. 

44.  I  can  pasture  10  horses  or  15  cows  on  my  ground;  if  I  have 
9  cows,  how  many  horses  can  I  keep?  Ans.  4. 

45.  A's  money  is  12  %  of  B's,  and  1G  %  of  C's;  B  has  $100  more 
than  C  :  how  much  has  A  ?  Ans.  §48. 

46.  Eight  men  hire  a  coach ;  by  getting  G  more  passengers,  the  ex- 
pense to  each  is  diminished  $l| :   what  do  they  pay  for  the  coach? 

.  Ans.  $32  f 

47.  A  company  engage  a  supper;  beingjoined  by  |  as  many  more, 
the  bill  of  eacli  is  00  ct.  less:  what  would  each  have  paid,  if  none 
had  joined  them?  Ans.  $i!.10 

48.  I!y  mixing  5  Ib.  of  good  sugar  with  31b.  worth  4  ct.  a  Ib.  less, 
the  mixture  is  worth  8^  ct.  a  Ib. :  find  the  prices  of  the  ingredients. 

Ans.  10 ct.  and  Get.  a  Ib. 

40.  Bj'  mixing  10 Ib.  of  good  sugar  with  Gib.  worth  only  J  as  much, 
the  mixture  is  worth  1  ct.  it  Ib.  less  than  the  good  sugar:  find  the 
prices  of  the  ingredients  and  of  the  mixture. 

Ans.  Ingredients,  8  ct.  and  5i  ct.;  Mixt.,  7  ct.  per  Ib. 
.50.  A  and  B  have  the  same  money ;  if  A  had  $20  more,  and  B,  $10 
less,  A  would  have  2i  times  as  much  as  B:  what  has  each? 

Ans.  $32  A 

ART.  471.          PRACTICAL  EXAMPLES. 

^^       1.   What  is  the  least  sum  that  can  be  paid  with  quarters,  dimes, 
or  3ct.  pieces  ?  Ans.  $1.00 

2.  A  and  1?  pa}'  SI. 75  for  a  quart  of  varnish,  and  10  ct.  for  the 
bottle;    A    contributes   $1,    B,    the    rest:    they  divide    the  varnish 
equally,  and  A  keeps  the  bottle  :    which  owes  the  other,  and  how 
much?  Ans.  B  owes  A  2A  ct. 

3.  Find  the  difference  between  3  sq.ft.  and  3  foot  square;  also  be 
"    tween  {  cu.  ft.  and  |  ft.  cubed.  Ans.  32  sq.  in.;  405  cu.  in. 

4.  How  far  does  a  man  walk,  while  planting  a  field  of  corn  285  ft 
square,  the  rows  being  3  ft.  apart  and  3  ft.  from  the  fences  ? 

Ans.  5  mi.  G  rd.  G  ft. 

5.  How  far  apart  should  the  knots  of  a  log-line  be,  to  indicate 
•very  half-minute  a  speed  of  1  mi.  per  hour  ?  Ans.  44  ft. 

6.  I  bought  25  sheep  for  $50  :    if  I  had  got  3  more  for  the  same 
Bam,  how  much  less  per  head  would  I  have  paid  ?  Ann.  24  ct. 

7.  Find  the  least  number  which,  divided  by  2,  3,  4,  ">,  and  •'>.  leaves 
a  remainder  1  each  time.  ,-l//.s'.  111. 

8.  How  many  carats  in  gold  coiuisting  of  1  r  wt.  lg  gr.  pure  gold 
md  .15  gr.  alloy  ?  Ann.  21  -,3/. . 


PROMISCUOUS   EXAMPLES.  367 

9.  Until  recently  the  pay  of  a  congressman  was  §8  a  clay  during 
the  session,  and  $8  for  every  20  mi.  he   traveled   in  going  to  anil 
coming  from  (lie  capital:  how  far  did  one  live  from  the  seat  of  gov- 
ernment, whose  pay  for  a  session  of  158  days,  w-.is  $'2104? 

Ans.'  1050  mi. 

10.  Land  worth  $1000  an  acre,  is  worth  how  much  a  foot  front  o? 
9U  ft.  depth  ;  reserving  Tlg  for  streets  ?  Ans.  $2.295+, 

11.  Reduce  f,  ?»,  |,  }°,  -$j  to  the  least  common  numerator. 

^ifimttMlfclfl 

12.  I  buy  stocks  at  20  %  discount,  and  sell  them  at  10  %  premiunr.: 
what  per  cent,  do  I  gain?  Ans.  37A  %. 

13.  I  invest  and  sell  at  a  loss  of  15  ^  ;  I  invest  the  proceeds  again, 
and  sell  at  a  gain  of  15  %  '•  do  I  gain  or  lose  on  the  two  speculations, 
and  how  many  per  cent.?  Ana.  Lose  2|  fc. 

14.  I  sell  at  8  %  gain,  invest  the  proceeds  and  sell  at  an  advance 
of  12A  <fe  ;  invest  the  proceeds  again,  and  sell  at  4  %  loss,  and  quit 
with  $1166.40  :   what  did  I  start  with  ?  Ans.  $1000. 

15.  I  can  insure  my  house  for  $2500,  at  y8g  %  premium  annually, 
or  permanently,  by  paying  down  12  annual  premiums  :  which  should 
I  prefer,  and  how  much  will  I  gain  by  it,  if  money  is  '.rorth  G  <&  per 
annum  to  me?  Any.  The  latter ;  gain,  $113.33!,- 

16.  A  owes  B  $1500,  due  in  1  yr.  10  mon.    He  pays  him  $300  cash, 
and  a  note  of  6  mon.  for  the  balance:  what  is  the  face  of  the  note, 
allowing  interest  at  6  %  ?  Ans.  $1080.56 

17.  If  I  charge  12  fe  per  annum  compound  interest,  parable  quar- 
terly, ^vhat  rate  per  annum  is  that?  Ans.  l^jVoVoVo^- 

18.  How  many  square  inches  in  one  face  of  a  cube  which  contains 
2571353 cu.  in.?  Ans.  18769. 

19.  What  is  the  side  of  a  cube  which  contains  as  many  cubic  inches, 
as  there  are  square  inches  in  its  surface?  Ans.  6  in. 

20.  If  12i  %  of  what  I  receive  for  an  article,  is  gain,  what  is  my 
rate  of  gain?  Ans.14'%%,. 

21.  The  boundaries  of  a  square  and  circle  are  each  20  ft.  ;  which 
contains  the  most,  and  how  much?    Ans.  Circle;  6.831  sq.  ft.  nearly. 

22.  If  I  pay  $1000  for  a  5yr.  lease  and  $200  for  repairs,  how  much 
rent  payable  quarterly  is  that  equal  to,  allowing  10  %  interest? 

Am.  $307.92  a  year. 

23.  What  is   the  value  of  a  widow's  dower  in  property  worth 
$3000,  her  age  being  40,  and  interest  5  %  1  Ans.  $669.50 

24.  Bought  eggs  on  credit;   the  first  time  one  dozen,  and  every 
succeeding  time  3  more.     My  last  purchase  was  7i  dozen.     The  bill 
was  presented  for  120  doz.:  how  much  too  large  was  it?    Ans.'&\doz. 

25.  What  principal  must  bo  loaned  Jan.  1,  at  9  ^,  to  be  re-paid  by 
5  installments  of  $200  each,  payable  on  the  first  day  of  the  five  suc- 
ceeding months?  Ans.  $978.10 

26.  After  spending  25  %  of  my  money,  and  25  j&  of  the  remain- 
der, I  had  left  $675  :  what  had  I  at  first?  Ans.  $1200. 

27.  I  had  a  60  day  note  discounted  at  1  fe  a  month  and  paid  S4.80 
above  true  interest:  what  was  the  facg  of  theno.e?  Ans.  $11112.93 


368  RAY'S   HIGHER   ARITHMETIC 


28.  If  the  draft  is  7  lb.,  the  tare  18  %,  and  the  net  weight  14cwt 
1  qr.  3  )!.>.,  what  is  the  prross  weight?  Ann.  17  cwt.  1  qr.  25  lb. 

2'.'.  Bought  slock  at  G  %  discount  and  sold  at  3  ^  premium,  ^fin- 
ing §180  :  what  did  I  invest?  Ans.  $-137.0!) 

30.  Bought  stock  at  20  f0  discount;  sold  out  at  45  <fi>  gain,  realiz 
lag  $2204:  what  did  I  invest?  -  Ans.  $1520. 

31.  Invested  $10000;  sold  out  at  a  loss  of  20  %  :  how  much  must 
I  borrow  at  4  %,  so  that  by  investing  all  1  have  at  18  ft,  i  may  re 
trieve  my  loss  ?  Ans.  $4000. 

32.  If  .{  of  an  inch  of  rain  falls,  how  many  bbl.  will  be  caught  by 
a  cistern  which  drains  a  roof  52  ft.  by  38  ft?          Ana.  9.770-}-  bbl. 

33.  The  tax  on  a  lot,  with  2  years'  penalty,  amounts  to  $275.924  : 
if  the  rate  of  taxation  for  the   1st  year  was  1-ffg  %,  and  for  the  2d  , 
year  lyVo  %}  and  the  penalty  is  30  %   of  the  tax;  what  is  the  lot 
assessed?  A ti.i.  $7500. 

34.  A  father  left  $20000  to  be  divided  among  his  4  sons,  aged  6 
years,  8  years,  10  years,  and  12  years,   respectively,  so    that   each  . 
i»liarc,  placed  at  4.>  %  compound  interest,  should  amount  to  the  sam« 
when  its  possessor  became  of  age  (21  yr.) :  what  were  the  shares  ? 

Ans.  $4300.34;  $4701.59;  $5199.78;  $5078.29 

35.  $30000  of  bonds  bearing  7  %  interest,  payable  semi-annually, 
and  due  in  20  yr.  are  bought  so  as  to  yield  8  ft,  payable  scmi-au  " 
nually  :  what  is  the  price?  Am.  $27031.08 

30.  I  sell  a  lot. for  $4850;  $250  payable  at  6 mon.,  Hyr.,  2iyr., 
3-  yr.,  and  4A  yr.  each,  and  the  balance  in  5yr.:  if  money  is  worth 
10  '%  per  annum  to  me,  what  is  the  cash  value  of  the  sale  ? 

Ans.  $3228.14 

37.  A,  B.  and  C  are  partners ;  A  puts  in  $240  for  G  mon.,  B  a  cer- 
tain sum  for  12  mon.,  C,  $100  for  a  certain  time;  when  they  close  up, 
A  has  $300,  B,  $000  ;  C,  $200:  find  B's  stock  and  C's  time. 

Ans.  $400  and  15  mon. 

38.  The  stocks  of  3  partners,  A,  B  and  C,  lire  in  trade  8,  10,  and 
7  mon.     respectively;    and    their    gains    are    $115.50,  $204.75    and 
£lS:;.7-~>  respectively:  find  their  stocks,  the  difference  between  B's 
and  C's  being  $220.  Ans.  A's,  $550;  B's,  $780;  C's,  $1000. 

39.  The  stocks  of  3  partners,  A,  B,  and  C,  arc  $350,  $220  and 
$250, and  their  gains  $112,  $88  and  $120  respectively:  find  the  time 
each  stock  was  in  trade,  B's  time  being  2  mon.  longer  than  A's. 

A»s.  A's,  8  mon.;  B's,  10  mon.;  C's,  12  mon. 

40.  By  discounting  a  note  at  20  </0   per  annum,  I  get  22^  %  per 
innum  interest:  how  long  does  the  note  run?  Ans.  200  days. 

41.  A'debt  is  to  be  paid  in  4  equal  installments  at  4,  9,  12,  20 
months  respectively;  its  cash  value  is    $750,  allowing  5   ^  simple 
interest:  what  is  the  debt?  Ans.  $7«4..4 

42.  I  bought  a  horse  for  $150  due  in  8  months,  and  sold  him  at 
once  for  $180:  find  the  gain  per  cent.;  int.  4 -A-  </c.       Ans.  1>  j  ',  '/r. 

43.  A  receives  $"i7. '.'(),  and   15.  $2!.'. 70,  from  a  jnint  speculation:  if 
L  invested  $7.83^  more  than  B,  what  did  each  invest  ? 

An.«.  A,  S1G.OSJ  ;  B,  $8.25 


PROMISCUOUS  EXAMPLES.  3611 

44.  A  borrows  a  sum  of  money  at  G  $,  payable  semi-annually, 
and  leads  it  at  12  ^,  payable  quarterly,  and  clears  $2450.85  a  year 
what  is  the  sum  ?  Am.  $38485.87 

45.  Find  the  sura,  whose  true  discount  by  simple  interest  for  4  yr, 
is  S'J5  more  at  6  %,  than  at  4  %  per  annum.  Am.  §449.50 

46.  I  invested  $2700  in  stock  at  25  %  discount,  which  pays  8  % 
annual  dividends:  how  much  must  I  invest  in  stock  at  4  </0  discoun 
and  paying  10  %  annual  dividends,  to  secure  an  equal  income? 

Ans.  $2704.80 

47.  Exchanged  $5200  01  rf-ack  bearing  5  f0  interest  at  C9  %,  fo 
stock  bearing  7  fo  interest  at  92  <&,  the  interest  on  each  stock  having 
been  just  paid :  what  is  my  cash  gain,  if  money  is  worth  0  <fc  to 
me  ?  Ans.  $216.66§ 

48.  Bought  goods  on  4mon.  credit;  after  7mon.  I  sell  them  for 
$1500,  2.T  %  off  for  cash;  my  gain  is  15  %,  money  being  worth  G  y'0: 
what  did"  I  pay  for  the  goods?  Ans.  $1252.94 

49.  The  8th  term  of  an  Arithmetical  series  is  76,  and  the  34th 
term,  200:  what  is  the  19th  term?  Ans.  131. 

50.  The  9th  term  of  a  geometric  series  is  137781,  and  the  13th 
term,  11160261:  what  is  the  4th  term?  Ans.  567. 

61.  My  capital  increases  every  year  by  the  same  per  cent.;  at  the 
end  of  the  3d  year,  it  was  $13310;  at  the  end  of  the  7th  year,  it  was 
$19487.171:  what  was  my  original  capital,  and  the  rate  of  gain? 

Ans.  $10000  and  10  %. 

52.  Find  the  length  of  a  minute-hand,  whose  extreme  point  moves 
4  inches  in  3  min.  28  sec.  Ans.  11.02 — in. 

53.  Three  men  own  a  grindstone,  2ft.  Sin.  in  diameter:  how  many 
inches  must  each  grind  off,  to  get  an  equal  share,  allowing  6  in.  waste 
for  the  aperture  ? 

Ans.  1st,  2.822— in.,  2d,  3.621-f-in.;  3d,  6.557— in. 

54.  A,  B  and  C  were  partners:  A  put  in  $600  for  3mon.;  B,  $400 
for  4mon.;  C,  $700:  the  gain  was  $120,  of  which  C  got  $35:  how 
long  was  C's  money  in  trado?  Ans.  "2  mon. 

55.  I  sold  an  article  at  20  %  g-iin;  had  it  cost  me  $300  more,  I 
would  have  lost  20  %:  find  the  cost.  Ans.  $600. 

66.  A  boat  goes  16  j  miles  an  hour  down   stream,  and  10  mi.  an 
hour  up  stream:  if  it  is  22  A  hr.  longer  in  coming  up  than  in  going 
down,  liow  far  down  did  it  go?  Ans.  585  mi. 

67.  If  an  article  had  cost  me  10  %  less,  my  rate  of  gain  would  have 
been  15  <fa  more:  what  was  my  rate  of  gain?  Ans.  35  %. 

58.  Bought  a  check  on  a  suspended  bank  nt  55  % ',  exchanged  it 
for  railroad  bonds  at  60  fo,  which  bear  1  %  interest;  what  rate  of 
iuterjst  do  I  receive  on  the  amount  of  money  invested? 

Ans.  21373  %. 

69.  Bought  sugar  for  refinery;  6  %  is  wasted  in  the  process;  30  * 
becomes  molasses,  which  is  sold  at  40  %  less  than  the  same  weight 
of  sugar  cost ;  at  what  %  advance  on  the  first  cost,  must  the  clarified 
•ugar  be  sold,  so  as  to  yield  a  profit  of  14  %  on  the  investment? 

Ant  60  %. 


370  RAY'S  HIGHER  ARITHMETIC. 


CP"  Many  questions  usually  solved  by  Algebra,  are  susceptillo 
of  easy  and  elegant  Arithmetical  solutions.  The  following  are 
from  'Ray's  Algebra.' 

GO.  A  servant  was  hired  for  1  yr.  at  £8  and  his  livery ;  in  7  mon. 
he  left,  find  received  £'2  13  s.  4  d.  and  his  livery  ;  what  was  the  value 
of  the  livery?  Ans.  £4  10  s. 

61.  A  smuggler  had  brandy  which  he  held  at  $198;  after  selling 
10  gal.,  a  revenue  officer  seized  |  of  the  remainder,  by  which  hi? 
receipts  on  the  whole  were  cut  down  tc  $162;  how  many  gallons  had 
he,  and  at  what  price?  Ans.  22  gal.  at  $9  a  gal. 

G2.  A  woi-kman  was  engaged  for  28  da.;  every  day  he  worked  he 
received  75  ct.,  and  every  day  he  was  absent  he  forfeited  25  ct.  At 
the  end  of  28  days  he  received  $12 :  how  many  days  did  he  work? 

Ans.  19. 

G3.  From  a  bag  of  money  which  contained  a  certain  sum.  there  was 
taken  $20  more  than  its  half;  from  the  remainder,  $30  more  than  its 
third;  and  from  the  remainder,  $40  more  than  its  fourth,  and  then 
nothing  was  left.  What  sum  did  it  contain?  .  Ans.  $290. 

G4.  A  bought  eggs  at  18  ct.  a  dozen  ;  had  he  bought  5  more  for  the 
same  sum,  they  would  have  cost  him  2^  ct.  a  dozen,  less.  How  many 
eggs  did  he  buy  1  Ans.  31. 

Go.  A  person  bought  sheep  for  $94;  having  lost  7,  he  sold  $  the 
remainder  at  prime  cost,  for  $20.  How  many  sheep  had  he  at  first? 

Ans.  47. 

G6.  For  every  8  sheep  a  farmer  keeps,  he  plows  1  A.  of  land,  and 
keeps  1  A.  of  pasture  for  every  5  sheep ;  how  many  sheep  does  he 
keep  on  325  A.  ?  Ans.  1000. 

67.  A  person  has  just  2  hr.  spare  time ;  how  far  may  he  ride  iu  a 
stage  which  travels  12  mi.  an  hour   so  as  to  return  home  in  time, 
walking  back  at  the  rate  of  4  mi.  an  hour?  Ans.  G  mi. 

68.  If  65  Ib.  of  sea-water  contain  2  Ib.  of  salt,   how  much   fresh 
water  must  be  added  to  these  G5  Ib.,  so  that  25  Ib.  of  the  new  mix- 
ture shall  contain  but  4  oz.  salt?  Ans.  135  Ib. 

69.  A  grocer  knows  neither  the  weight  nor  first  cost  of  a  box  of 
tea.     He  only  recollects  that  if  he  had  sold  the  whole  at  30  ct.  a  Ib., 
he  would  have  gained  $1,   but  if  he  had  sold  it  at  22  ct.  a  Ib.,  he 
would  have  lost  $3.    Find  the  number  of  Ib.  in  the  box,  and  the  first 
cost  per  Ib.  Ans.  50  Ib.  at  28  ct. 

70.  A,  B  and  C  are  paid  $10  74  for  doing  a  piece  of  work;  A  and 
B  together  could  do  £  of  it,  in  the  same  time  that  A  and  C  do  -jj  of 
it,  or  B  and  C  do  f  of  it;  how  must  they  divide  the  money? 

Ans.  A  gets  $7.02  ;  B,  $5.94 ;  C,  $3.78 

THE    END. 


Just  Published. — A  KEY,  embracing  full  and  lucid  Solutions  to 
the  Examples  in  '  Ray's  Higher  Arithmetic ' 


ECLECTIC    SYSTEM 

OF 

PENMANSHIP. 


ECLECTIC  COPY-BOOKS. 

No.  1 ;  No.  2 ;  No.  3 ;  No.  4 ;  No.  6  Boys ;  No.  6  Girls ; 

No.  5  Boys ;  No.  5  Girls ;  No.  7,  Book  of  Forms;  No.  7  Girls; 

No.  8,  Variety  of  Capitals;  No.  8  Girls,  (nearly  ready;) 
No.  9,  Book  of  Off-hand  Writing,  German  and  old  English  Test,  etc., 
(nearly  ready.) 

035"  Tfie  Girls'  booJts  contain  the  snine  copies  as  the  corresponding 
inimlters  for  Roys,  but  in  smaller  handwriting — a  feature  peculiar  to 
the  Eclectic  System  of  Penmanship.  ~53i 

The  simplest,  most  legible,  and  business-like  style  of  capitals  and  small 
letters  is  adopted,  in  the  conviction  that  this  standard  will  be  the  most 
acceptable  to  teachers,  and  valuable  to  students.  Each  letter  is  given 
separately  at  first,  and  then  in  combination.  The  spacing  is  open,  mak- 
ing the  writing  legible,  and  easily  written. 

Thf.  analysis  is  simple,  and  indicated  in  every  letter  when  first  presented. 
Explanation*  are  clear,  concise,  and  complete,  and  are  given  on  the 
cover  of  each  book,  and  not  over  and  around  the  copy,  to  distract  at- 
tention. 


ECLECTIC  EXERCISE-BOOK. 

THE  EXERCISE- BOOK  con  tains  a  variety  of  exercises  especially  designed 
to  develop  the  different  movements,  and  so  arranged  as  to  give  as  much 
or  as  little  practice  on  eacli  exercise  as  may  be  desired,  without  the  aid  of 
extra  paper.  It  is  a  little  larger  than  the  Copy-Books,  and  has  a  strong 
cover,  in  order  that  the  Copy-Book  may  be  placed  within  it;  thus  mak- 
ing it  convenient  to  keep  the  two  together.  Every  pupil  should  be  supplied 
with  it. 


ECLECTIC  HAND-BOOK  OF  PENMANSHIP. 

TUK  HAND-BOOK  contains  the  true  plan  of  instruction  ;  a  complete  de- 
scription and  analysis  of  movement  and  of  the  letters  ;  lessons  on  form, 
shading,  and  spacing;  a  chapter  011  teaching  writing  in  ungraded 
schools,  the  "  neglected  art,"  etc. ;  and  the  substance  of  what  is  required 
in  teaching  Penmanship.  Copious  illustrations  and  diagrams  are  given, 
showing  the  forms  of  letters,  and  manner  of  instruction. 


ECLECTIC  WRITING-CARDS; 

Another  New  Feature. 

THE  ECLECTIC  WRITING-CARDS  comprise  thirty -six  Cards,  9x13  inches, 
with  loop  attached  for  suspending  along  the  walls  of  the  school-room. 

The  first,  twenty-six  Cards  present  each  a  capital  letter  on  one  side,  and 
n  small  letter  on  the  reverse.  Figures,  curves,  etc.,  are  represented  on 
the  remainder  of  the  Cards. 

The  letters  are  white  on  a  black  ground,  and  are  of  a  larger  size  than  is 
usually  employed  in  Writing  Charts.  They  can  be  easily  read  and 
studied  at  a  distance  of  thirty  to  fifty  feet. 

For  convenience  in  instructing  classes,  a  description  of  each  letter  is 
given  on  its  Card.  It  is  confidently  believed  that  Teachers  will  find  these 
Cards  by  far  the  most  Heauti/ul,  Useful,  and  Practical  Aids  in  teaching  this 
important  branch  of  education  yet  published. 

The  thirty-six  Cards  are  put  up  in  a  neat  box,  for  convenience  in  hand- 
ling, and  lor  their  bolter  preservation  when  not  in  use. 


SCHOOL     BOOKS. 

ECLECTIC    EDUCATIONAL    SERIES. 


McGnffey's  Scries. 

.Eclectic  Speller. 
New  First  Reader. 
New  Second  Reader. 
New  Third  Reader. 
New  Fourth  Reader. 
New  Fifth  Reader. 
New  Sixth  Reader. 
Primary  Charts,  (10  Nos.) 

Ray's  Arithmetics. 

Primary  Arithmetic. 
Intellectual  Arithmetic. 
Rudiments  of  Arithmetic. 
Practical  Arithmetic. 
Higher  Arithmetic. 
Test  Examples. 

Ray's  Algebras. 

New  Elementary  Algebra. 
New  Higher  Algebra. 

Ray's  Higher  Mathematics. 

Plane  and  Solid  Geometry. 
Geometry  and  Trigonometry. 
Analytic  Geometry. 
Elements  of  Astronomy. 
Calculus,  fin  preparation.) 
Surveying  &  Navigation,  (in  prep.) 
Evans's  School  Geometry. 

Grammar  and  Composition. 

HARVEY'S  Elementary  Gram. 
HARVEY'S  English  Grammar. 
PINNEO'S  Primary  Grammar. 
PINNEO'S  Analytical  Grammar. 
PINNEO'S  Guide  to  Composit'n. 
PINNEO'S  English  Teacher. 
PINNEO'S  False  Syntax. 
PINNEO'S  Parsing  Exercises. 


The  Eclectic  Geographies. 

Primary  Geography. 
Intermediate  Geography. 
School  Geography. 

The  Eclectic  Penmanship. 

Copy-Books,  (12  Nos.) 
Writing-Cards,  (36.) 
Hand-Book  of  Penmanship. 
Exercise-Book. 

School  Music. 

PHILLIPS'  Day  School  Singer. 
The  Young  Singer,  No.  I. 
The  Young  Singer,  No.  II. 
The  Young  Singer's  Manual. 

Logic,  Philosophy,  etc. 

SCHUYLER'S  Logic. 
NORTON'S  Natural  Philosophy. 
KIDD'S  Elocution. 
McGUFFEY'S  Primary  Speaker. 
McGUFFEY'S  Eclectic  Speaker. 
COLE'S  Institute  Reader. 

Graded  School  Series. 

"WHITE'S  Primary  Arithmetic. 
WHITE'S  Intermediate  Arith. 
WHITE'S  Complete  Arithmetic, 
SCHUYLER'S  Comp.  Algebra. 

Registers. 

WHITE'S  Com.  School  Register. 
WHITE'S  Grad.  School  Register. 

Miscellaneous. 

DeWOLF'S  Instructive  Speller. 
LEIGH'S  Phonetic  Primer. 
LEIGH'S  Phonetic  Reader. 
The  Examiner,  or  Teacher's  Aid. 


WILSON,  HINKLE  &  CO., 


SS  Bond   Street. 


CINCINNATI. 


NEW  YOSE. 


This  book  is  DUE  on  the  last  date  stamped  below 


? 

•APR  1  1  1932 
JAN  23  1933 

FEB  2      1933 

FEB  1  7  1933 


MM  4  - 


FEB  6 
MAY  1  6 


Form  L-9-35m-8,'28 


UC  SOUTHERN  REGIONAL  LIBRARY  FACILITY 


001  204612    4 


UNiV 


AT. 

ANG 
JBRARY 


